Given sn = 1 + q + q² + .... + qⁿ; Sn =...

MathematicsProperties of Binomial CoefficientsJEE Advanced 1984Moderate

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

Visualized Solution (Hindi)

Understanding the terms .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s_{n} and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S_{n}

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ (A G.P. with $n + 1$ terms)
Sum of G.P. formula: $s_{k} = \frac{1 - q ^{k + 1}}{1 - q}$
Given $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}$
This is also a G.P. with $n + 1$ terms and common ratio $\frac{q + 1}{2}$

Expressing LHS in Summation Form

LHS $=^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n}$
In Sigma notation: LHS $= \sum_{k = 0}^{n}^{n + 1} C_{k + 1} s_{k}$

Substituting the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s_{k}

Substitute $s_{k} = \frac{1 - q ^{k + 1}}{1 - q}$ into the LHS expression:
LHS $= \sum_{k = 0}^{n}^{n + 1} C_{k + 1} (\frac{1 - q ^{k + 1}}{1 - q})$
LHS $= \frac{1}{1 - q} [\sum_{k = 0}^{n}^{n + 1} C_{k + 1} - \sum_{k = 0}^{n}^{n + 1} C_{k + 1} q^{k + 1}]$

Applying Binomial Identities

Let $r = k + 1$ . As $k$ goes from $0$ to $n$ , $r$ goes from $1$ to $n + 1$ .
First sum: $\sum_{r = 1}^{n + 1}^{n + 1} C_{r} = (2^{n + 1} -^{n + 1} C_{0}) = 2^{n + 1} - 1$
Second sum: $\sum_{r = 1}^{n + 1}^{n + 1} C_{r} q^{r} = ((1 + q)^{n + 1} -^{n + 1} C_{0} q^{0}) = (1 + q)^{n + 1} - 1$

Simplifying the LHS Result

LHS $= \frac{1}{1 - q} [(2^{n + 1} - 1) - ((1 + q)^{n + 1} - 1)]$
LHS $= \frac{2 ^{n + 1} - ( 1 + q ) ^{n + 1}}{1 - q}$

Evaluating the RHS

RHS $= 2^{n} S_{n} = 2^{n} \sum_{k = 0}^{n} (\frac{q + 1}{2})^{k}$
Using G.P. sum formula: RHS $= 2^{n} [\frac{1 - ( \frac{q + 1}{2} ) ^{n + 1}}{1 - \frac{q + 1}{2}}]$

Simplifying the RHS Denominator

Simplify the denominator: $1 - \frac{q + 1}{2} = \frac{2 - ( q + 1 )}{2} = \frac{1 - q}{2}$
Substitute back: RHS $= 2^{n} \cdot \frac{1 - ( \frac{q + 1}{2} ) ^{n + 1}}{\frac{1 - q}{2}}$
RHS $= 2^{n} \cdot 2 \cdot \frac{1 - \frac{( q + 1 ) ^{n + 1}}{2 ^{n + 1}}}{1 - q}$

Final Comparison and Proof

RHS $= 2^{n + 1} \cdot \frac{\frac{2 ^{n + 1} - ( q + 1 ) ^{n + 1}}{2 ^{n + 1}}}{1 - q}$
The $2^{n + 1}$ terms cancel out:
RHS $= \frac{2 ^{n + 1} - ( q + 1 ) ^{n + 1}}{1 - q}$
LHS = RHS. Hence Proved.

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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 1983Easy

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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 $2^{2 n - 1} - \frac{( 2 n )!}{2 ( n ! ) ^{2}}$

MathematicsProperties of Binomial CoefficientsJEE Advanced 1989Moderate

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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 Advanced 1994Moderate

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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 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 $p < q$ .

MathematicsSum of Special SeriesJEE Advanced 1988Easy

View in:English Hindi

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

(A)

2^{n} - n - 1

(B)

1 - 2^{- n}

(C)

n + 2^{- n} - 1

(D)

2^{n} + 1

Answer:C

MathematicsProperties of Binomial CoefficientsJEE Advanced 2003Moderate

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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 1999Moderate

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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 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 Main 2004Easy

View in:English Hindi

If $S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}}$ and $t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}$ , then $\frac{t _{n}}{S _{n}}$ is equal to

(A)

\frac{2 n - 1}{2}

(B)

\frac{n}{2} - 1

(C)

n - 1

(D)

\frac{n}{2}

Answer:D

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

Visualized Solution (Hindi)

Understanding the terms $s_{n}$ and $S_{n}$

Expressing LHS in Summation Form

Substituting the value of $s_{k}$

Applying Binomial Identities

Simplifying the LHS Result

Evaluating the RHS

Simplifying the RHS Denominator

Final Comparison and Proof

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

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 $2^{2 n - 1} - \frac{( 2 n )!}{2 ( n ! ) ^{2}}$

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

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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 $p < q$ .

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

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

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 $m \leq n$ .

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 $S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}}$ and $t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}$ , then $\frac{t _{n}}{S _{n}}$ is equal to

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

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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 $2^{2 n - 1} - \frac{( 2 n )!}{2 ( n ! ) ^{2}}$

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

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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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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 $m \leq n$ .

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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​

Visualized Solution (Hindi)

Understanding the terms .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sn​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Expressing LHS in Summation Form

Substituting the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sk​

Applying Binomial Identities

Simplifying the LHS Result

Evaluating the RHS

Simplifying the RHS Denominator

Final Comparison and Proof

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

Understanding the terms $s_{n}$ and $S_{n}$

Substituting the value of $s_{k}$

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

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

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 $S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}}$ and $t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}$ , then $\frac{t _{n}}{S _{n}}$ is equal to

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

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

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 $p < q$ .

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

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

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 $S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}}$ and $t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}$ , then $\frac{t _{n}}{S _{n}}$ is equal to