If f(m, n) = (1-x^m)(1-x^ m-1 )...(1-x^...

MathematicsProperties of Binomial CoefficientsJEE Advanced 1978Moderate

If $f (m, n) = \frac{( 1 - x ^{m} ) ( 1 - x ^{m - 1} ) \dots ( 1 - x ^{m - n + 1} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n} )}$ where $m$ and $n$ are positive integers ( $n \leq m$ ), show that $f (m, n + 1) = f (m - 1, n + 1) + x^{m - n - 1} f (m - 1, n)$ .

Visualized Solution (Hindi)

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (m, n)

Given function: $f (m, n) = \frac{( 1 - x ^{m} ) ( 1 - x ^{m - 1} ) \dots ( 1 - x ^{m - n + 1} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n} )}$
This is known as the Gaussian Binomial Coefficient, often denoted as $(n m)_{x}$ .
Constraint: $m, n \in Z^{+}$ and $n \leq m$ .
Objective: Prove $f (m, n + 1) = f (m - 1, n + 1) + x^{m - n - 1} f (m - 1, n)$ .

Defining the Target Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (m, n + 1)

Substitute $n \to n + 1$ in the general formula:
$f (m, n + 1) = \frac{( 1 - x ^{m} ) ( 1 - x ^{m - 1} ) \dots ( 1 - x ^{m - (n + 1) + 1} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n + 1} )}$
Simplifying the last term in the numerator:
$f (m, n + 1) = \frac{( 1 - x ^{m} ) ( 1 - x ^{m - 1} ) \dots ( 1 - x ^{m - n} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n + 1} )}$

Expanding the RHS Terms

Term 1: $f (m - 1, n + 1) = \frac{( 1 - x ^{m - 1} ) ( 1 - x ^{m - 2} ) \dots ( 1 - x ^{m - n - 1} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n + 1} )}$
Term 2: $f (m - 1, n) = \frac{( 1 - x ^{m - 1} ) ( 1 - x ^{m - 2} ) \dots ( 1 - x ^{m - n} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n} )}$

Finding the Common Factor

Observe the relationship: $f (m - 1, n + 1) = f (m - 1, n) \cdot \frac{1 - x ^{m - n - 1}}{1 - x ^{n + 1}}$
Now, write the full RHS expression:
RHS $= f (m - 1, n) \cdot \frac{1 - x ^{m - n - 1}}{1 - x ^{n + 1}} + x^{m - n - 1} f (m - 1, n)$
Factor out $f (m - 1, n)$ :
RHS $= f (m - 1, n) [\frac{1 - x ^{m - n - 1}}{1 - x ^{n + 1}} + x^{m - n - 1}]$

Simplifying the Bracket

Simplify the term inside the bracket by taking LCM:
$\frac{1 - x ^{m - n - 1} + x ^{m - n - 1} ( 1 - x ^{n + 1} )}{1 - x ^{n + 1}}$
Expand the numerator:
$\frac{1 - x ^{m - n - 1} + x ^{m - n - 1} - x ^{(m - n - 1) + (n + 1)}}{1 - x ^{n + 1}}$
Cancel the terms $- x^{m - n - 1}$ and $+ x^{m - n - 1}$ :
$\frac{1 - x ^{m}}{1 - x ^{n + 1}}$

Final Synthesis

Combine the result with the common factor:
RHS $= f (m - 1, n) \cdot \frac{1 - x ^{m}}{1 - x ^{n + 1}}$
Substitute the definition of $f (m - 1, n)$ :
RHS $= \frac{( 1 - x ^{m - 1} ) \dots ( 1 - x ^{m - n} )}{( 1 - x ) \dots ( 1 - x ^{n} )} \cdot \frac{1 - x ^{m}}{1 - x ^{n + 1}}$
Rearranging the terms:
RHS $= \frac{( 1 - x ^{m} ) ( 1 - x ^{m - 1} ) \dots ( 1 - x ^{m - n} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n + 1} )} = f (m, n + 1)$
LHS = RHS

Key Takeaways and Next Challenges

Key Takeaway: The identity $(n + 1 m)_{x} = (n + 1 m - 1)_{x} + x^{m - n - 1} (n m - 1)_{x}$ is the $q$ -analog of the Pascal identity.
As $x \to 1$ , $f (m, n) \to (n m)$ , and the identity becomes $(n + 1 m) = (n + 1 m - 1) + (n m - 1)$ .
Next Challenge: Try to prove the dual identity: $f (m, n + 1) = x^{n + 1} f (m - 1, n + 1) + f (m - 1, n)$ .

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Conceptually Similar Problems

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Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

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

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(A)

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Answer:D

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(A)

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(B)

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(D)

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If the expansion in powers of $x$ of the function $\frac{1}{( 1 - a x ) ( 1 - b x )}$ is $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} ....$ then $a_{n}$ is

(A)

\frac{b ^{n} - a ^{n}}{b - a}

(B)

\frac{a ^{n} - b ^{n}}{b - a}

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(D)

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Answer:D

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

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

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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 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 $b_{n} =^{2 n + 1} C_{n + 1}$ .

If $f (m, n) = \frac{( 1 - x ^{m} ) ( 1 - x ^{m - 1} ) \dots ( 1 - x ^{m - n + 1} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n} )}$ where $m$ and $n$ are positive integers ( $n \leq m$ ), show that $f (m, n + 1) = f (m - 1, n + 1) + x^{m - n - 1} f (m - 1, n)$ .

Visualized Solution (Hindi)

Understanding the Function $f (m, n)$

Defining the Target Term $f (m, n + 1)$

Expanding the RHS Terms

Finding the Common Factor

Simplifying the Bracket

Final Synthesis

Key Takeaways and Next Challenges

Conceptually Similar Problems

Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

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

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 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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If $f (m, n) = \frac{( 1 - x ^{m} ) ( 1 - x ^{m - 1} ) \dots ( 1 - x ^{m - n + 1} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n} )}$ where $m$ and $n$ are positive integers ( $n \leq m$ ), show that $f (m, n + 1) = f (m - 1, n + 1) + x^{m - n - 1} f (m - 1, n)$ .

Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

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

If $f (x) = x^{n}$ , then the value of $f (1) - \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} - \frac{f ^{'''} ( 1 )}{3 !} + \dots + \frac{( - 1 ) ^{n} f ^{n} ( 1 )}{n !}$ is

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If the expansion in powers of $x$ of the function $\frac{1}{( 1 - a x ) ( 1 - b x )}$ is $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} ....$ then $a_{n}$ is

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

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

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

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 $b_{n} =^{2 n + 1} C_{n + 1}$ .

Visualized Solution (Hindi)

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(m,n)

Defining the Target Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(m,n+1)

Expanding the RHS Terms

Finding the Common Factor

Simplifying the Bracket

Final Synthesis

Key Takeaways and Next Challenges

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Understanding the Function $f (m, n)$

Defining the Target Term $f (m, n + 1)$

Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

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If $f (x) = x^{n}$ , then the value of $f (1) - \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} - \frac{f ^{'''} ( 1 )}{3 !} + \dots + \frac{( - 1 ) ^{n} f ^{n} ( 1 )}{n !}$ is

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Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

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

If $f (x) = x^{n}$ , then the value of $f (1) - \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} - \frac{f ^{'''} ( 1 )}{3 !} + \dots + \frac{( - 1 ) ^{n} f ^{n} ( 1 )}{n !}$ is

Let $f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}$ for $n \geq 2$ and $g (x) = (f \circ f \circ ... \circ f) (x)$ ( $f$ occurs $n$ times). Then $\int x^{n - 2} g (x) d x$ equals

If the expansion in powers of $x$ of the function $\frac{1}{( 1 - a x ) ( 1 - b x )}$ is $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} ....$ then $a_{n}$ is

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

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 $b_{n} =^{2 n + 1} C_{n + 1}$ .