Prove that C₀ - 2² C₁ + 3² C₂ - .... +...

MathematicsProperties of Binomial CoefficientsJEE Advanced 1989Moderate

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

Visualized Solution (English)

Identifying the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T_{r}

Let the given sum be $S$ .
The general term of the series is $T_{r} = (- 1)^{r} (r + 1)^{2} C_{r}$ , where $r = 0, 1, 2, \dots, n$ .
The complete sum is $S = \sum_{r = 0}^{n} (- 1)^{r} (r + 1)^{2} C_{r}$ .

Expanding the Quadratic Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (r + 1)^{2}

Expand the quadratic part: $(r + 1)^{2} = r^{2} + 2 r + 1$ .
Substitute this into the general term: $T_{r} = (- 1)^{r} (r^{2} + 2 r + 1) C_{r}$ .

Using Falling Factorials for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r^{2}

To simplify the summation, rewrite $r^{2}$ as $r (r - 1) + r$ .
Thus, $(r + 1)^{2} = [r (r - 1) + r] + 2 r + 1 = r (r - 1) + 3 r + 1$ .
The general term becomes: $T_{r} = (- 1)^{r} [r (r - 1) + 3 r + 1] C_{r}$ .

Splitting the Summation

Using the linearity of summation:
$S = \sum_{r = 0}^{n} (- 1)^{r} r (r - 1) C_{r} + 3 \sum_{r = 0}^{n} (- 1)^{r} r C_{r} + \sum_{r = 0}^{n} (- 1)^{r} C_{r}$ .

Property of Binomial Coefficients

Recall the property: $r C_{r} = n \cdot^{n - 1} C_{r - 1}$ .
Applying it twice: $r (r - 1) C_{r} = n (n - 1) \cdot^{n - 2} C_{r - 2}$ .

Evaluating the First Sum

First term: $n (n - 1) \sum_{r = 2}^{n} (- 1)^{r}^{n - 2} C_{r - 2}$ .
Let $k = r - 2$ , then the sum is $n (n - 1) \sum_{k = 0}^{n - 2} (- 1)^{k + 2}^{n - 2} C_{k}$ .
Since $(- 1)^{k + 2} = (- 1)^{k}$ , the sum is $n (n - 1) (1 - 1)^{n - 2} = 0$ for $n > 2$ .

Evaluating the Second Sum

Second term: $3 n \sum_{r = 1}^{n} (- 1)^{r}^{n - 1} C_{r - 1}$ .
This is $3 n \sum_{k = 0}^{n - 1} (- 1)^{k + 1}^{n - 1} C_{k} = - 3 n (1 - 1)^{n - 1} = 0$ for $n > 1$ .

Evaluating the Third Sum

Third term: $\sum_{r = 0}^{n} (- 1)^{r} C_{r}$ .
This is the standard expansion of $(1 - 1)^{n} = 0$ for $n > 0$ .

Combining the Results

Combining all parts:
$S = 0 + 3 (0) + 0 = 0$ .
This result holds true for all $n > 2$ .

Conclusion and Key Takeaways

Key Takeaway: Use $r (r - 1) \dots (r - k + 1)^{n} C_{r} = n (n - 1) \dots (n - k + 1)^{n - k} C_{r - k}$ to reduce terms.
Condition Check: The sum is zero because each component represents $(1 - 1)^{m}$ where $m \geq 1$ .
Next Challenge: Try proving the identity for $\sum (- 1)^{r} (r + 1)^{3} C_{r}$ .

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 $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

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Prove that $\sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0$ , where $k = (3 n) /2$ and $n$ is an even positive integer.

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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 $p < q$ .

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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 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

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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 1986Moderate

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If $C_{r}$ stands for $^{n} C_{r}$ , then the sum of the series $\frac{2 ( \frac{n}{2} )! ( \frac{n}{2} )!}{n !} [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - .... + (- 1)^{n} (n + 1) C_{n}^{2}]$ , where $n$ is an even positive integer, is equal to

(A)

(B)

(- 1)^{n /2} (n + 1)

(C)

(- 1)^{n /2} (n + 2)

(D)

(- 1)^{n} n

Answer:C

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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 $m \leq n$ .

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 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 $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

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

Visualized Solution (English)

Identifying the General Term $T_{r}$

Expanding the Quadratic Term $(r + 1)^{2}$

Using Falling Factorials for $r^{2}$

Splitting the Summation

Property of Binomial Coefficients

Evaluating the First Sum

Evaluating the Second Sum

Evaluating the Third Sum

Combining the Results

Conclusion and 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 $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

Prove that $\sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0$ , where $k = (3 n) /2$ and $n$ is an even positive integer.

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 $p < q$ .

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

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 $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .

If $C_{r}$ stands for $^{n} C_{r}$ , then the sum of the series $\frac{2 ( \frac{n}{2} )! ( \frac{n}{2} )!}{n !} [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - .... + (- 1)^{n} (n + 1) C_{n}^{2}]$ , where $n$ is an even positive integer, is equal to

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

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 $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

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

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 $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

Prove that $\sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0$ , where $k = (3 n) /2$ and $n$ is an even positive integer.

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 $p < q$ .

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

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 $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .

If $C_{r}$ stands for $^{n} C_{r}$ , then the sum of the series $\frac{2 ( \frac{n}{2} )! ( \frac{n}{2} )!}{n !} [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - .... + (- 1)^{n} (n + 1) C_{n}^{2}]$ , where $n$ is an even positive integer, is equal to

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

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 $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

.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​.

Visualized Solution (English)

Identifying the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr​

Expanding the Quadratic Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (r+1)2

Using Falling Factorials for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2

Splitting the Summation

Property of Binomial Coefficients

Evaluating the First Sum

Evaluating the Second Sum

Evaluating the Third Sum

Combining the Results

Conclusion and Key Takeaways

Conceptually Similar Problems

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.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; } 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; } 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​.

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.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; } Prove that \sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0, where k=(3n)/2 and n is an even positive integer.

.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; } 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; } 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; } 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; } 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; } 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​.

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

Identifying the General Term $T_{r}$

Expanding the Quadratic Term $(r + 1)^{2}$

Using Falling Factorials for $r^{2}$

Prove that $\sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0$ , where $k = (3 n) /2$ and $n$ is an even positive integer.

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 $p < q$ .

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

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 $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

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

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 $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

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

Prove that $\sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0$ , where $k = (3 n) /2$ and $n$ is an even positive integer.

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 $p < q$ .

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

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 $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

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

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 $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .