JEE Advanced 1992

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: 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 $b_{n} =^{2 n + 1} C_{n + 1}$ .

Q: Why does the first part of the sum $\sum_{r=0}^{n-1} a_r(y+1)^r$ vanish?

Because the highest power of $y$ in $(y+1)^r$ is $r$. If $r$ only goes up to $n-1$, the highest power is $y^{n-1}$. It simply lacks the 'strength' to reach $y^n$. Since we are looking for the coefficient of $y^n$, this entire segment contributes exactly zero.

Q: Why do we look for the coefficient of $y^{n+1}$ in the numerator?

Because our sum formula resulted in a division by $y$. If we have an expression $\frac{P(y)}{y}$, the coefficient of $y^n$ in the whole expression is the coefficient of $y^{n+1}$ in $P(y)$. It is a simple index shift caused by the division.

Progressive Hints

Hint 1

The core of this problem lies in the mismatch of the polynomial centers. We are comparing a polynomial centered at

x = 2

with one centered at

x = 3

. To compare coefficients, we must bring them to the same 'home ground'. Can you think of a substitution that aligns these two?

Hint 2

Once you substitute

y = x - 3

, the right side becomes a simple power series in

y

. The left side becomes a summation of

(y + 1)^{r}

. Now, the problem reduces to finding the coefficient of

y^{n}

in this new expression. Recall that for a polynomial identity, the coefficients of corresponding powers must be equal.

Hint 3

You will find that the summation splits into two parts. The first part (where

r < n

) cannot contribute to the

y^{n}

term. Focus your energy on the second part:

\sum_{r = n}^{2 n} (y + 1)^{r}

. Recognize this as a Geometric Progression and use the sum formula

S = A \frac{R ^{N} - 1}{R - 1}

to simplify it before extracting the coefficient.

Visualized Solution

The Given Equation

We are given: $\sum_{r = 0}^{2 n} a_{r} (x - 2)^{r} = \sum_{r = 0}^{2 n} b_{r} (x - 3)^{r}$
We also know that $a_{k} = 1$ for all $k \geq n$ .
Our goal is to find the value of $b_{n}$ .

Substitution $y = x - 3$

To find $b_{n}$ , we need to isolate the terms involving $b_{r}$ .
The right side has $(x - 3)^{r}$ . Let's simplify this by substituting $y = x - 3$ .
This implies $x = y + 3$ , so $x - 2 = y + 1$ .

Transforming to Variable $y$

Substitute $y = x - 3$ and $x - 2 = y + 1$ into the original equation.
Left Hand Side (LHS) becomes: $\sum_{r = 0}^{2 n} a_{r} (y + 1)^{r}$
Right Hand Side (RHS) becomes: $\sum_{r = 0}^{2 n} b_{r} y^{r}$
New Equation: $\sum_{r = 0}^{2 n} a_{r} (y + 1)^{r} = \sum_{r = 0}^{2 n} b_{r} y^{r}$

Expanding $\sum_{r = 0}^{2 n} b_{r} y^{r}$

Let's expand the RHS: $\sum_{r = 0}^{2 n} b_{r} y^{r}$
$\sum_{r = 0}^{2 n} b_{r} y^{r} = b_{0} + b_{1} y + b_{2} y^{2} + \dots + b_{n} y^{n} + \dots + b_{2 n} y^{2 n}$
Notice that $b_{n}$ is simply the coefficient of $y^{n}$ in this expansion.

Equating Coefficients for $b_{n}$

Since LHS = RHS, the coefficient of $y^{n}$ must be the same on both sides.
Therefore, $b_{n} = coefficient of y^{n} in the LHS$ .
$b_{n} = coefficient of y^{n} in \sum_{r = 0}^{2 n} a_{r} (y + 1)^{r}$

Splitting the Summation at $r = n$

We know $a_{k} = 1$ for $k \geq n$ .
Let's split the LHS summation at $r = n$ .
$\sum_{r = 0}^{2 n} a_{r} (y + 1)^{r} = \sum_{r = 0}^{n - 1} a_{r} (y + 1)^{r} + \sum_{r = n}^{2 n} a_{r} (y + 1)^{r}$

Applying Condition $a_{k} = 1$

In the second sum, $r \geq n$ , so we can replace $a_{r}$ with $1$ .
$\sum_{r = 0}^{2 n} a_{r} (y + 1)^{r} = \sum_{r = 0}^{n - 1} a_{r} (y + 1)^{r} + \sum_{r = n}^{2 n} 1 \cdot (y + 1)^{r}$
Now we need the coefficient of $y^{n}$ in this entire expression.

Analyzing $\sum_{r = 0}^{n - 1} a_{r} (y + 1)^{r}$

Consider the first part: $\sum_{r = 0}^{n - 1} a_{r} (y + 1)^{r}$
The maximum power of $(y + 1)$ in this sum is $n - 1$ .
Therefore, the highest power of $y$ generated by this part is $y^{n - 1}$ .
It contains no $y^{n}$ term.

Focusing on $\sum_{r = n}^{2 n} (y + 1)^{r}$

Since the first part contributes nothing to $y^{n}$ , we only look at the second part.
$b_{n} = coefficient of y^{n} in \sum_{r = n}^{2 n} (y + 1)^{r}$
Let's write out this series: $(y + 1)^{n} + (y + 1)^{n + 1} + \dots + (y + 1)^{2 n}$

Recognizing the G.P.

The series $(y + 1)^{n} + (y + 1)^{n + 1} + \dots + (y + 1)^{2 n}$ is a Geometric Progression (G.P.).
First term, $A = (y + 1)^{n}$
Common ratio, $R = (y + 1)$
Number of terms, $N = (2 n - n) + 1 = n + 1$

Sum of the G.P.

The sum of a G.P. is $S = A \frac{R ^{N} - 1}{R - 1}$
Substitute the values: $S = (y + 1)^{n} \frac{( y + 1 ) ^{n + 1} - 1}{( y + 1 ) - 1}$
Simplify the denominator: $(y + 1) - 1 = y$
$S = \frac{( y + 1 ) ^{n} [( y + 1 ) ^{n + 1} - 1 ]}{y}$

Simplifying to $\frac{( y + 1 ) ^{2 n + 1} - ( y + 1 ) ^{n}}{y}$

Multiply $(y + 1)^{n}$ inside the bracket:
Numerator becomes: $(y + 1)^{2 n + 1} - (y + 1)^{n}$
The complete sum is: $S = \frac{( y + 1 ) ^{2 n + 1} - ( y + 1 ) ^{n}}{y}$
We need the coefficient of $y^{n}$ in this expression.

Targeting Coefficient of $y^{n + 1}$

We want the coefficient of $y^{n}$ in $\frac{( y + 1 ) ^{2 n + 1} - ( y + 1 ) ^{n}}{y}$
Because of the division by $y$ , finding the coefficient of $y^{n}$ in the whole expression is equivalent to finding the coefficient of $y^{n + 1}$ in the numerator.
Target: Coefficient of $y^{n + 1}$ in $[(y + 1)^{2 n + 1} - (y + 1)^{n}]$

Evaluating Final Coefficient $^{2 n + 1} C_{n + 1}$

Numerator has two parts: $(y + 1)^{2 n + 1}$ and $(y + 1)^{n}$
In $(y + 1)^{n}$ , the highest power is $y^{n}$ . So, the coefficient of $y^{n + 1}$ is $0$ .
In $(y + 1)^{2 n + 1}$ , the coefficient of $y^{n + 1}$ is given by the binomial theorem as $^{2 n + 1} C_{n + 1}$ .
Therefore, $b_{n} =^{2 n + 1} C_{n + 1} - 0 =^{2 n + 1} C_{n + 1}$ .

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The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

We are given the identity:

r = 0 \sum 2 n a_{r} (x - 2)^{r} = r = 0 \sum 2 n b_{r} (x - 3)^{r}

We are tasked with finding the value of

b_{n}

, given the constraint that

a_{k} = 1

for all

k \geq n

The Change of Variables

To compare the coefficients, we must align the centers of the polynomials. Let

y = x - 3

This substitution implies

x = y + 3

, which leads to

x - 2 = y + 1

. The identity now transforms into:

r = 0 \sum 2 n a_{r} (y + 1)^{r} = r = 0 \sum 2 n b_{r} y^{r}

We seek

b_{n}

, which is the coefficient of

y^{n}

in the expansion of the left-hand side.

Splitting the Summation

Since

a_{k} = 1

for

k \geq n

, we split the left-hand summation into two parts:

r = 0 \sum n - 1 a_{r} (y + 1)^{r} + r = n \sum 2 n (y + 1)^{r}

The first part,

\sum_{r = 0}^{n - 1} a_{r} (y + 1)^{r}

, is a polynomial of degree at most

n - 1

. Consequently, it cannot contribute to the coefficient of

y^{n}

Evaluating the Geometric Progression

We focus our attention on the second part:

S = r = n \sum 2 n (y + 1)^{r}

This is a geometric progression with first term

A = (y + 1)^{n}

, common ratio

R = (y + 1)

, and

n + 1

terms. Using the sum formula

S = A \frac{R ^{n + 1} - 1}{R - 1}

, we obtain:

S = (y + 1)^{n} \frac{( y + 1 ) ^{n + 1} - 1}{( y + 1 ) - 1} = \frac{( y + 1 ) ^{2 n + 1} - ( y + 1 ) ^{n}}{y}

Final Calculation

To find the coefficient of

y^{n}

S

, we look for the coefficient of

y^{n + 1}

in the numerator:

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

The term

(y + 1)^{n}

has a maximum degree of

n

and contributes nothing to the

y^{n + 1}

coefficient. Applying the Binomial Theorem to

(y + 1)^{2 n + 1}

, the coefficient of

y^{n + 1}

is given by:

(n + 1 2 n + 1)

Thus, the final value is $b_{n} = (n + 1 2 n + 1)$ .

Common Traps & FAQs

Why can't we just compare coefficients of

(x - 2)^{r}

and

(x - 3)^{r}

directly?

Because coefficients are only comparable when the basis is the same. It is like comparing apples and oranges. By shifting the origin with

y = x - 3

, we transform both sides into polynomials of

y

, effectively making them both 'apples'. Only then can we equate the coefficients of

y^{n}

Why does the first part of the sum

\sum_{r = 0}^{n - 1} a_{r} (y + 1)^{r}

vanish?

Because the highest power of

y

(y + 1)^{r}

r

. If

r

only goes up to

n - 1

, the highest power is

y^{n - 1}

. It simply lacks the 'strength' to reach

y^{n}

. Since we are looking for the coefficient of

y^{n}

, this entire segment contributes exactly zero.

Why do we look for the coefficient of

y^{n + 1}

in the numerator?

Because our sum formula resulted in a division by

y

. If we have an expression

\frac{P ( y )}{y}

, the coefficient of

y^{n}

in the whole expression is the coefficient of

y^{n + 1}

P (y)

. It is a simple index shift caused by the division.

Animated Solution for Mathematics - Binomial Theorem: 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

The Given Equation

Substitution $y = x - 3$

Transforming to Variable $y$

Expanding $\sum_{r = 0}^{2 n} b_{r} y^{r}$

Equating Coefficients for $b_{n}$

Splitting the Summation at $r = n$

Applying Condition $a_{k} = 1$

Analyzing $\sum_{r = 0}^{n - 1} a_{r} (y + 1)^{r}$

Focusing on $\sum_{r = n}^{2 n} (y + 1)^{r}$

Recognizing the G.P.

Sum of the G.P.

Simplifying to $\frac{( y + 1 ) ^{2 n + 1} - ( y + 1 ) ^{n}}{y}$

Targeting Coefficient of $y^{n + 1}$

Evaluating Final Coefficient $^{2 n + 1} C_{n + 1}$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Change of Variables

Splitting the Summation

Evaluating the Geometric Progression

Final Calculation

Similar Questions

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

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

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

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

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

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

Animated Solution for Mathematics - Binomial Theorem: .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​.

Visualized Solution

The Given Equation

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x−3

Transforming to Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑r=02n​br​yr

Equating Coefficients for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​

Splitting the Summation at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=n

Applying Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ak​=1

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑r=0n−1​ar​(y+1)r

Focusing on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑r=n2n​(y+1)r

Recognizing the G.P.

Sum of the G.P.

Simplifying to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(y+1)2n+1−(y+1)n​

Targeting Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } yn+1

Evaluating Final Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2n+1Cn+1​

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Change of Variables

Splitting the Summation

Evaluating the Geometric Progression

Final Calculation

Similar Questions

.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; } 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; } 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; } 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; } 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 a positive integer and (1+x+x2)n=a0​+a1​x+a2​x2+....+a2n​x2n. Show that a02​−a12​+a22​−....+a2n2​=an​.

Animated Solution for Mathematics - Binomial Theorem: 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 $b_{n} =^{2 n + 1} C_{n + 1}$ .

Substitution $y = x - 3$

Transforming to Variable $y$

Expanding $\sum_{r = 0}^{2 n} b_{r} y^{r}$

Equating Coefficients for $b_{n}$

Splitting the Summation at $r = n$

Applying Condition $a_{k} = 1$

Analyzing $\sum_{r = 0}^{n - 1} a_{r} (y + 1)^{r}$

Focusing on $\sum_{r = n}^{2 n} (y + 1)^{r}$

Simplifying to $\frac{( y + 1 ) ^{2 n + 1} - ( y + 1 ) ^{n}}{y}$

Targeting Coefficient of $y^{n + 1}$

Evaluating Final Coefficient $^{2 n + 1} C_{n + 1}$

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

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.

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 $m \leq n$ .

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

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

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