If (1 + x)ⁿ = C₀ + C₁x + C₂x² + .... +...

MathematicsProperties of Binomial CoefficientsJEE Advanced 1983Easy

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + .... + C_{n} x^{n}$ then show that the sum of the products of the $C_{i}$ 's taken two at a time, represented by $\sum_{0 \leq i < j \leq n} C_{i} C_{j}$ is equal to $2^{2 n - 1} - \frac{( 2 n )!}{2 ( n ! ) ^{2}}$

Visualized Solution (Hindi)

The Binomial Expansion

Given expansion: $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + ... + C_{n} x^{n}$
Objective: Find the sum of products taken two at a time: $\sum_{0 \leq i < j \leq n} C_{i} C_{j}$

The Square of a Sum Identity

Recall the algebraic identity:
$(\sum_{i = 0}^{n} a_{i})^{2} = \sum_{i = 0}^{n} a_{i}^{2} + 2 \sum_{0 \leq i < j \leq n} a_{i} a_{j}$
Applying this to binomial coefficients $C_{i}$ :
$(\sum_{i = 0}^{n} C_{i})^{2} = \sum_{i = 0}^{n} C_{i}^{2} + 2 \sum_{0 \leq i < j \leq n} C_{i} C_{j}$

Sum of Coefficients .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum C_{i}

To find $\sum C_{i}$ , substitute $x = 1$ in $(1 + x)^{n}$ :
$(1 + 1)^{n} = C_{0} + C_{1} + C_{2} + ... + C_{n}$
$\sum_{i = 0}^{n} C_{i} = 2^{n}$
Therefore, $(\sum C_{i})^{2} = (2^{n})^{2} = 2^{2 n}$

Sum of Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum C_{i}^{2}

Standard result for sum of squares of coefficients:
$\sum_{i = 0}^{n} C_{i}^{2} = (n 2 n)$
In factorial form:
$\sum_{i = 0}^{n} C_{i}^{2} = \frac{( 2 n )!}{( n ! ) ^{2}}$

Substituting into the Identity

Substitute values into: $(\sum C_{i})^{2} = \sum C_{i}^{2} + 2 \sum C_{i} C_{j}$
$2^{2 n} = \frac{( 2 n )!}{( n ! ) ^{2}} + 2 \sum_{0 \leq i < j \leq n} C_{i} C_{j}$

Isolating the Product Term

Rearrange to solve for $2 \sum C_{i} C_{j}$ :
$2 \sum_{0 \leq i < j \leq n} C_{i} C_{j} = 2^{2 n} - \frac{( 2 n )!}{( n ! ) ^{2}}$

The Final Result

Divide by $2$ :
$\sum_{0 \leq i < j \leq n} C_{i} C_{j} = \frac{2 ^{2 n}}{2} - \frac{( 2 n )!}{2 ( n ! ) ^{2}}$
$\sum_{0 \leq i < j \leq n} C_{i} C_{j} = 2^{2 n - 1} - \frac{( 2 n )!}{2 ( n ! ) ^{2}}$
Hence Proved.

Key Takeaways

Key Concept: Use $(\sum a_{i})^{2}$ to find $\sum_{i < j} a_{i} a_{j}$ .
Standard Sums: $\sum C_{i} = 2^{n}$ and $\sum C_{i}^{2} = (n 2 n)$ .
Next Challenge: Try finding $\sum_{0 \leq i < j \leq n} (- 1)^{i + j} C_{i} C_{j}$ .

00:00 / 00:00

Conceptually Similar Problems

MathematicsProperties of Binomial CoefficientsJEE Advanced 1979Moderate

View in:English Hindi

Given that $C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + .... + 2 n C_{2 n} x^{2 n - 1} = 2 n (1 + x)^{2 n - 1}$ where $C_{r} = \frac{( 2 n )!}{r ! ( 2 n - r )!}, r = 0, 1, 2, ...., 2 n$ . Prove that $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

MathematicsProperties of Binomial CoefficientsJEE Advanced 1994Moderate

View in:English Hindi

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

MathematicsProperties of Binomial CoefficientsJEE Advanced 1984Moderate

View in:English Hindi

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

MathematicsProperties of Binomial CoefficientsJEE Advanced 2000Moderate

View in:English Hindi

For any positive integer $m, n$ (with $n \geq m$ ), let $(m n) =^{n} C_{m}$ . Prove that $(m n) + (m n - 1) + (m n - 2) + .... + (m m) = (m + 1 n + 1)$ . Hence or otherwise, prove that $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .

MathematicsProperties of Binomial CoefficientsJEE Advanced 2003Moderate

View in:English Hindi

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

MathematicsProperties of Binomial CoefficientsJEE Advanced 1989Moderate

View in:English Hindi

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

MathematicsProperties of Binomial CoefficientsJEE Advanced 1992Moderate

View in:English Hindi

If $\sum_{r = 0}^{2 n} a_{r} (x - 2)^{r} = \sum_{r = 0}^{2 n} b_{r} (x - 3)^{r}$ and $a_{k} = 1$ for all $k \geq n$ , then show that $b_{n} =^{2 n + 1} C_{n + 1}$ .

MathematicsProperties of Binomial CoefficientsJEE Advanced 1989Moderate

View in:English Hindi

Prove that $C_{0} - 2^{2} C_{1} + 3^{2} C_{2} - .... + (- 1)^{n} (n + 1)^{2} C_{n} = 0, n > 2$ , where $C_{r} =^{n} C_{r}$ .

MathematicsProperties of Binomial CoefficientsJEE Main 2008Easy

View in:English Hindi

Statement-1: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1}$ . Statement-2: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1}$ .

(A)

Statement-1 is false, Statement-2 is true

(B)

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

(C)

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

(D)

Statement-1 is true, Statement-2 is false

Answer:B

MathematicsProperties of Binomial CoefficientsJEE Advanced 1999Moderate

View in:English Hindi

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + .... + C_{n} x^{n}$ then show that the sum of the products of the $C_{i}$ 's taken two at a time, represented by $\sum_{0 \leq i < j \leq n} C_{i} C_{j}$ is equal to $2^{2 n - 1} - \frac{( 2 n )!}{2 ( n ! ) ^{2}}$

Visualized Solution (Hindi)

The Binomial Expansion

The Square of a Sum Identity

Sum of Coefficients $\sum C_{i}$

Sum of Squares $\sum C_{i}^{2}$

Substituting into the Identity

Isolating the Product Term

The Final Result

Key Takeaways

Conceptually Similar Problems

Given that $C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + .... + 2 n C_{2 n} x^{2 n - 1} = 2 n (1 + x)^{2 n - 1}$ where $C_{r} = \frac{( 2 n )!}{r ! ( 2 n - r )!}, r = 0, 1, 2, ...., 2 n$ . Prove that $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

For any positive integer $m, n$ (with $n \geq m$ ), let $(m n) =^{n} C_{m}$ . Prove that $(m n) + (m n - 1) + (m n - 2) + .... + (m m) = (m + 1 n + 1)$ . Hence or otherwise, prove that $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

If $\sum_{r = 0}^{2 n} a_{r} (x - 2)^{r} = \sum_{r = 0}^{2 n} b_{r} (x - 3)^{r}$ and $a_{k} = 1$ for all $k \geq n$ , then show that $b_{n} =^{2 n + 1} C_{n + 1}$ .

Prove that $C_{0} - 2^{2} C_{1} + 3^{2} C_{2} - .... + (- 1)^{n} (n + 1)^{2} C_{n} = 0, n > 2$ , where $C_{r} =^{n} C_{r}$ .

Statement-1: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1}$ . Statement-2: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1}$ .

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + .... + C_{n} x^{n}$ then show that the sum of the products of the $C_{i}$ 's taken two at a time, represented by $\sum_{0 \leq i < j \leq n} C_{i} C_{j}$ is equal to $2^{2 n - 1} - \frac{( 2 n )!}{2 ( n ! ) ^{2}}$

Given that $C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + .... + 2 n C_{2 n} x^{2 n - 1} = 2 n (1 + x)^{2 n - 1}$ where $C_{r} = \frac{( 2 n )!}{r ! ( 2 n - r )!}, r = 0, 1, 2, ...., 2 n$ . Prove that $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

For any positive integer $m, n$ (with $n \geq m$ ), let $(m n) =^{n} C_{m}$ . Prove that $(m n) + (m n - 1) + (m n - 2) + .... + (m m) = (m + 1 n + 1)$ . Hence or otherwise, prove that $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

If $\sum_{r = 0}^{2 n} a_{r} (x - 2)^{r} = \sum_{r = 0}^{2 n} b_{r} (x - 3)^{r}$ and $a_{k} = 1$ for all $k \geq n$ , then show that $b_{n} =^{2 n + 1} C_{n + 1}$ .

Prove that $C_{0} - 2^{2} C_{1} + 3^{2} C_{2} - .... + (- 1)^{n} (n + 1)^{2} C_{n} = 0, n > 2$ , where $C_{r} =^{n} C_{r}$ .

Statement-1: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1}$ . Statement-2: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1}$ .

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

Visualized Solution (Hindi)

The Binomial Expansion

The Square of a Sum Identity

Sum of Coefficients .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑Ci​

Sum of Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑Ci2​

Substituting into the Identity

Isolating the Product Term

The Final Result

Key Takeaways

Conceptually Similar Problems

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be a positive integer and (1+x+x2)n=a0​+a1​x+a2​x2+....+a2n​x2n. Show that a02​−a12​+a22​−....+a2n2​=an​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given sn​=1+q+q2+....+qn; Sn​=1+2q+1​+(2q+1​)2+....+(2q+1​)n,q=1 Prove that n+1C1​+n+1C2​s1​+n+1C3​s2​+....+n+1Cn+1​sn​=2nSn​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that 2k(0n​)(kn​)−2k−1(1n​)(k−1n−1​)+2k−2(2n​)(k−2n−2​)−....+(−1)k(kn​)(0n−k​)=(kn​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using mathematical induction, prove that mC0n​Ck​+mC1n​Ck−1​+....+mCkn​C0​=(m+n)Ck​, where m,n,k are positive integers, and pCq​=0 for p<q.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑r=02n​ar​(x−2)r=∑r=02n​br​(x−3)r and ak​=1 for all k≥n, then show that bn​=2n+1Cn+1​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that C0​−22C1​+32C2​−....+(−1)n(n+1)2Cn​=0,n>2, where Cr​=nCr​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement-1: ∑r=0n​(r+1)nCr​=(n+2)2n−1. Statement-2: ∑r=0n​(r+1)nCr​xr=(1+x)n+nx(1+x)n−1.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be any positive integer. Prove that ∑k=0m​(n2n−k​)(k2n−k​)​(2n−4k+1)2n−2k=(n−m2n−2m​)(mn​)​2n−2m for each non-negative integer m≤n.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be a positive integer and (1+x+x2)n=a0​+a1​x+a2​x2+....+a2n​x2n. Show that a02​−a12​+a22​−....+a2n2​=an​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given sn​=1+q+q2+....+qn; Sn​=1+2q+1​+(2q+1​)2+....+(2q+1​)n,q=1 Prove that n+1C1​+n+1C2​s1​+n+1C3​s2​+....+n+1Cn+1​sn​=2nSn​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that 2k(0n​)(kn​)−2k−1(1n​)(k−1n−1​)+2k−2(2n​)(k−2n−2​)−....+(−1)k(kn​)(0n−k​)=(kn​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using mathematical induction, prove that mC0n​Ck​+mC1n​Ck−1​+....+mCkn​C0​=(m+n)Ck​, where m,n,k are positive integers, and pCq​=0 for p<q.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑r=02n​ar​(x−2)r=∑r=02n​br​(x−3)r and ak​=1 for all k≥n, then show that bn​=2n+1Cn+1​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that C0​−22C1​+32C2​−....+(−1)n(n+1)2Cn​=0,n>2, where Cr​=nCr​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement-1: ∑r=0n​(r+1)nCr​=(n+2)2n−1. Statement-2: ∑r=0n​(r+1)nCr​xr=(1+x)n+nx(1+x)n−1.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be any positive integer. Prove that ∑k=0m​(n2n−k​)(k2n−k​)​(2n−4k+1)2n−2k=(n−m2n−2m​)(mn​)​2n−2m for each non-negative integer m≤n.

Sum of Coefficients $\sum C_{i}$

Sum of Squares $\sum C_{i}^{2}$

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

If $\sum_{r = 0}^{2 n} a_{r} (x - 2)^{r} = \sum_{r = 0}^{2 n} b_{r} (x - 3)^{r}$ and $a_{k} = 1$ for all $k \geq n$ , then show that $b_{n} =^{2 n + 1} C_{n + 1}$ .

Prove that $C_{0} - 2^{2} C_{1} + 3^{2} C_{2} - .... + (- 1)^{n} (n + 1)^{2} C_{n} = 0, n > 2$ , where $C_{r} =^{n} C_{r}$ .

Statement-1: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1}$ . Statement-2: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1}$ .

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

If $\sum_{r = 0}^{2 n} a_{r} (x - 2)^{r} = \sum_{r = 0}^{2 n} b_{r} (x - 3)^{r}$ and $a_{k} = 1$ for all $k \geq n$ , then show that $b_{n} =^{2 n + 1} C_{n + 1}$ .

Prove that $C_{0} - 2^{2} C_{1} + 3^{2} C_{2} - .... + (- 1)^{n} (n + 1)^{2} C_{n} = 0, n > 2$ , where $C_{r} =^{n} C_{r}$ .

Statement-1: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1}$ . Statement-2: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1}$ .

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .