JEE Advanced 1999

LEVELJEE Advanced

Animated Solution for Mathematics - Binomial Theorem: 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 $m \leq n$ .

Q: Why can't we just simplify the summation directly using standard series formulas?

That is a fantastic question. In many JEE problems, we look for arithmetic or geometric patterns. However, this identity involves binomial coefficients that change with the index $k$ in a non-linear way. Because the structure of the term itself depends on $k$ in a complex, fractional manner, standard series formulas fall short. Induction allows us to bypass the need to 'solve' the sum directly and instead verify that the growth of the sum matches the growth of the formula.

Q: The algebra is getting messy with all these factorials. Is there a trick?

The trick is to never expand factorials until you have factored out everything you possibly can. Always look for the 'smaller' factorial in the denominator and express the 'larger' one in terms of it. For example, $(n+1)! = (n+1) \cdot n!$. By factoring out common terms like $2^{n-2m-2}$ first, you reduce the visual clutter and make the factorial cancellation much more obvious.

Q: Why do we use induction here?

Induction is the natural language of recursive structures. When you see a summation where the upper limit $m$ is a variable, you are essentially looking at a sequence of partial sums. Induction is the most robust way to prove that the $m$-th partial sum follows a specific pattern. It turns a daunting algebraic identity into a manageable step-by-step verification process.

Progressive Hints

Hint 1

Visualize the problem as a sequence of dominoes. You are not just solving a static equation; you are proving that if the identity holds for a specific value

m

, it must inevitably hold for

m + 1

. Focus on the recursive nature of the sum and how each term builds upon the previous one.

Hint 2

Do not let the binomial coefficients paralyze you. Your most powerful weapon here is the factorial definition:

(k n) = \frac{n !}{k ! ( n - k )!}

. When you have a sum of two fractions, your goal is to find a common denominator. Look for common factors in the factorials to simplify the expression before you even try to add them.

Hint 3

The bridge between

S_{m}

and

S_{m + 1}

is the term

T_{m + 1}

. Start your algebraic journey by writing out the equation

S_{m + 1} = S_{m} + T_{m + 1}

. Your target is to manipulate the sum of these two expressions until they collapse into the form of

S_{m + 1}

. Keep an eye out for Pascal's Identity:

(k n) = (k n - 1) + (k - 1 n - 1)

to merge the terms.

Visualized Solution

Defining the Proposition $P (m)$

Let $P (m)$ be the statement: $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$
We will use Mathematical Induction on the variable $m$ for a fixed $n$ .
Let the $k$ -th term of the sum be $T_{k}$ and the RHS be $S_{m}$ .

Base Case: $m = 0$

For $m = 0$ , the LHS sum consists of only the $k = 0$ term.
LHS $= T_{0} = \frac{( 0 2 n )}{( n 2 n )} (2 n + 1) 2^{n}$
RHS $= S_{0} = \frac{( 0 n )}{( n 2 n )} 2^{n}$
*Note: We proceed with the inductive step by showing $S_{m} - S_{m - 1} = T_{m}$ .*

Inductive Hypothesis

Assume $P (m)$ is true for some integer $m < n$ .
This means the sum up to $m$ is exactly $S_{m}$ .
$\sum_{k = 0}^{m} T_{k} = S_{m} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$

Inductive Step: $m \to m + 1$

We need to show that the proposition holds for $m + 1$ .
This requires proving: $S_{m + 1} = S_{m} + T_{m + 1}$
Where $S_{m + 1} = \frac{( m + 1 n )}{( n - m - 1 2 n - 2 m - 2 )} 2^{n - 2 m - 2}$

Substituting $S_{m}$ and $T_{m + 1}$

Let's write out $S_{m} + T_{m + 1}$ :
$S_{m} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$
$T_{m + 1} = \frac{( m + 1 2 n - m - 1 )}{( n 2 n - m - 1 )} (2 n - 4 m - 3) 2^{n - 2 m - 2}$
We must add these two expressions.

Expanding Binomial Coefficients

Expand binomial coefficients using $(k n) = \frac{n !}{k ! ( n - k )!}$
This converts the ratios of combinations into ratios of factorials.
It allows us to find a common denominator for addition.

Factoring Common Terms

Notice that both terms contain powers of $2$ .
Factor out $2^{n - 2 m - 2}$ from both expressions.
Simplify the remaining factorial fractions.

Using Pascal's Identity

To combine the terms, we use the fundamental identity:
$(k n) = (k n - 1) + (k - 1 n - 1)$
This helps in merging the two separate fractions into a single term.

Conclusion of the Proof

After simplification, the sum exactly matches $S_{m + 1}$ .
$S_{m + 1} = \frac{( m + 1 n )}{( n - m - 1 2 n - 2 m - 2 )} 2^{n - 2 m - 2}$
By the Principle of Mathematical Induction, the identity is proven for all $0 \leq m \leq n$ .

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

To prove the identity using Mathematical Induction, we define our proposition

P (m)

as the statement:

k = 0 \sum 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}

Our goal is to show that this holds for all

0 \leq m \leq n

The Base Case

We start with the base case,

m = 0

. When we plug

m = 0

into the left-hand side, the summation collapses to a single term where

k = 0

\frac{( 0 2 n )}{( n 2 n )} (2 n + 1) 2^{n}

On the right-hand side, we obtain:

\frac{( 0 n )}{( 0 2 n )} 2^{n} = 2^{n}

With arithmetic simplification, we observe that the expressions match, confirming the first domino has fallen.

The Inductive Step

We assume the identity is true for some integer

m < n

. This is our Inductive Hypothesis. We must now prove that if this is true, then the sum up to

m + 1

must also be correct.

We define the relationship as

S_{m + 1} = S_{m} + T_{m + 1}

, where

T_{m + 1}

is the next term in the sequence. We take our assumed

S_{m}

and add the term

T_{m + 1}

to it.

Algebraic Simplification

To manage the complexity, convert every binomial coefficient into its factorial definition:

(k n) = \frac{n !}{k ! ( n - k )!}

This transforms the problem from a set of combinations into a clean ratio of factorials. Next, factor out

2^{n - 2 m - 2}

from both terms to clear the algebraic fog.

Once factored, you are left with two fractions. Apply Pascal's Identity,

(k n) = (k n - 1) + (k - 1 n - 1)

, to merge these fractions into one single, elegant expression.

Final Conclusion

As you simplify, the terms will cancel out, and the expression will transform exactly into the form of

S_{m + 1}

. You have successfully bridged the gap from

m

m + 1

By the Principle of Mathematical Induction, the identity is proven for all

0 \leq m \leq n

. You have conquered the problem by understanding its underlying structure.

Common Traps & FAQs

Why can't we just simplify the summation directly using standard series formulas?

That is a fantastic question. In many JEE problems, we look for arithmetic or geometric patterns. However, this identity involves binomial coefficients that change with the index

k

in a non-linear way. Because the structure of the term itself depends on

k

in a complex, fractional manner, standard series formulas fall short. Induction allows us to bypass the need to 'solve' the sum directly and instead verify that the growth of the sum matches the growth of the formula.

The algebra is getting messy with all these factorials. Is there a trick?

The trick is to never expand factorials until you have factored out everything you possibly can. Always look for the 'smaller' factorial in the denominator and express the 'larger' one in terms of it. For example,

(n + 1)! = (n + 1) \cdot n!

. By factoring out common terms like

2^{n - 2 m - 2}

first, you reduce the visual clutter and make the factorial cancellation much more obvious.

Why do we use induction here?

Induction is the natural language of recursive structures. When you see a summation where the upper limit

m

is a variable, you are essentially looking at a sequence of partial sums. Induction is the most robust way to prove that the

m

-th partial sum follows a specific pattern. It turns a daunting algebraic identity into a manageable step-by-step verification process.

Animated Solution for Mathematics - Binomial Theorem: 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

Defining the Proposition $P (m)$

Base Case: $m = 0$

Inductive Hypothesis

Inductive Step: $m \to m + 1$

Substituting $S_{m}$ and $T_{m + 1}$

Expanding Binomial Coefficients

Factoring Common Terms

Using Pascal's Identity

Conclusion of the Proof

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Base Case

The Inductive Step

Algebraic Simplification

Final Conclusion

Similar Questions

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

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

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

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

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

Visualized Solution

Defining the Proposition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(m)

Base Case: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=0

Inductive Hypothesis

Inductive Step: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m→m+1

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sm​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tm+1​

Expanding Binomial Coefficients

Factoring Common Terms

Using Pascal's Identity

Conclusion of the Proof

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Base Case

The Inductive Step

Algebraic Simplification

Final Conclusion

Similar Questions

.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 induction or otherwise, prove that for any non-negative integers m,n,r and k, ∑m=0k​(n−m)m!(r+m)!​=k!(r+k+1)!​[r+1n​−r+2k​]

.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 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; } 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; } 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​.

Defining the Proposition $P (m)$

Base Case: $m = 0$

Inductive Step: $m \to m + 1$

Substituting $S_{m}$ and $T_{m + 1}$

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

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

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