JEE Advanced 1994

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: 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 $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

Progressive Hints

Hint 1

The Spark: Look closely at the structure of the alternating sum

a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots + a_{2 n}^{2}

. This pattern strongly suggests the product of two polynomials, where one has alternating signs. How can you transform the original expansion

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

to introduce these alternating signs?

Hint 2

The Arsenal: Consider the substitution

x \to - \frac{1}{x}

. When you apply this to the original identity

(1 + x + x^{2})^{n} = \sum a_{r} x^{r}

, you create a new, related identity. How does this help you generate the alternating signs needed for the sum of squares?

Hint 3

The Execution: Multiply the original identity

(1 + x + x^{2})^{n} = \sum a_{r} x^{r}

by the modified identity

(1 - \frac{1}{x} + \frac{1}{x ^{2}})^{n} = \sum a_{r} (- 1)^{r} x^{- r}

after multiplying by

x^{2 n}

. Then, compare the coefficient of

x^{2 n}

on both sides of the resulting equation.

Visualized Solution

Given Multinomial Expansion

Given expansion: $(1 + x + x^{2})^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r}$
This can be written as: $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}$

The Substitution Trick: $x \to - \frac{1}{x}$

Replace $x$ with $- \frac{1}{x}$ in the identity.
$(1 - \frac{1}{x} + \frac{1}{x ^{2}})^{n} = \sum_{r = 0}^{2 n} a_{r} (- \frac{1}{x})^{r}$

Simplifying the Substituted Form

LHS: $(\frac{x ^{2} - x + 1}{x ^{2}})^{n} = \frac{( x ^{2} - x + 1 ) ^{n}}{x ^{2 n}}$
RHS: $\sum_{r = 0}^{2 n} a_{r} (- 1)^{r} x^{- r}$

Rearranging the Identity

Multiply both sides by $x^{2 n}$ :
$(x^{2} - x + 1)^{n} = \sum_{r = 0}^{2 n} a_{r} (- 1)^{r} x^{2 n - r}$

Multiplying the Two Expansions

Multiply Eq (1) and Eq (2):
$(1 + x + x^{2})^{n} (1 - x + x^{2})^{n} = (\sum_{r = 0}^{2 n} a_{r} x^{r}) (\sum_{k = 0}^{2 n} a_{k} (- 1)^{k} x^{2 n - k})$

Simplifying the LHS Product

LHS: $[(1 + x^{2}) + x]^{n} [(1 + x^{2}) - x]^{n} = [(1 + x^{2})^{2} - x^{2}]^{n}$
$= (1 + 2 x^{2} + x^{4} - x^{2})^{n} = (1 + x^{2} + x^{4})^{n}$

Extracting Coefficient from RHS

Coefficient of $x^{2 n}$ in RHS product occurs when $r = k$ .
Coeff of $x^{2 n} = \sum_{r = 0}^{2 n} a_{r} \cdot a_{r} (- 1)^{r} = a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots + a_{2 n}^{2}$

Extracting Coefficient from LHS

LHS is $(1 + x^{2} + (x^{2})^{2})^{n}$ .
Since $(1 + x + x^{2})^{n} = \sum a_{r} x^{r}$ , then $(1 + x^{2} + x^{4})^{n} = \sum a_{r} (x^{2})^{r} = \sum a_{r} x^{2 r}$ .
Coefficient of $x^{2 n}$ in this expansion is $a_{n}$ (at $r = n$ ).

Final Conclusion

Equating coefficients of $x^{2 n}$ from both sides:
$a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots + a_{2 n}^{2} = a_{n}$
Hence Proved.

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The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

We begin with the fundamental multinomial expansion:

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

This identity defines the sequence of coefficients

a_{0}, a_{1}, \dots, a_{2 n}

. Our objective is to prove the identity:

a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots + a_{2 n}^{2} = a_{n}

Phase 1

The Transformation

To generate the alternating signs required for the proof, we apply the substitution

x \to - \frac{1}{x}

. Substituting this into our original identity yields:

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

By taking a common denominator of

x^{2}

inside the bracket, we simplify the left side:

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

Multiplying both sides by

x^{2 n}

, we obtain a powerful secondary identity:

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

Phase 2

The Collision

We now multiply the original expansion by our modified expansion. The product of the left-hand sides is:

(1 + x + x^{2})^{n} (1 - x + x^{2})^{n} = [(1 + x^{2}) + x]^{n} [(1 + x^{2}) - x]^{n}

Recognizing the difference of squares pattern

(A + B) (A - B) = A^{2} - B^{2}

, this simplifies to:

[(1 + x^{2})^{2} - x^{2}]^{n} = (1 + 2 x^{2} + x^{4} - x^{2})^{n} = (1 + x^{2} + x^{4})^{n}

Phase 3

The Revelation

Next, we examine the product of the two series on the right-hand side:

(r = 0 \sum 2 n a_{r} x^{r}) (k = 0 \sum 2 n a_{k} (- 1)^{k} x^{2 n - k})

The coefficient of

x^{2 n}

in this product is obtained when

r = k

. This results in the sum:

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

Finally, we consider the left-hand side

(1 + x^{2} + x^{4})^{n}

. This is equivalent to the original expansion

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

where

y = x^{2}

The coefficient of

x^{2 n}

(1 + x^{2} + x^{4})^{n}

is identical to the coefficient of

x^{n}

in the original expansion, which is

a_{n}

. By equating the coefficients of

x^{2 n}

from both sides, we conclude:

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

Common Traps & FAQs

Why did we choose the substitution

x \to - \frac{1}{x}

? It seems completely arbitrary.

It is a brilliant, classic JEE-level trick! We need to generate an alternating sum of squares (

a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots

). By replacing

x

with

- 1/ x

, we introduce the negative sign needed for the alternating series and the reciprocal power

x^{- r}

that allows us to align the coefficients when we multiply the two series together. It is designed to make the powers of

x

cancel out perfectly in the product.

How do we know the coefficient of

x^{2 n}

is the one we need?

Great observation. When we multiply the two series, we get terms like

a_{r} x^{r}

and

a_{k} (- 1)^{k} x^{2 n - k}

. The product of these terms is

a_{r} a_{k} (- 1)^{k} x^{2 n + r - k}

. For this to be a term in

x^{2 n}

, the exponent

2 n + r - k

must equal

2 n

, which implies

r = k

. Thus, the coefficient of

x^{2 n}

is exactly the sum where

r = k

, which is

\sum a_{r}^{2} (- 1)^{r}

, giving us the alternating sum of squares.

Is it always valid to equate coefficients like this?

Yes, absolutely! If two polynomials are identically equal for all values of

x

, then the coefficients of the corresponding powers of

x

must be equal. Since we have derived an identity that holds for all

x

, we are mathematically justified in extracting and equating the coefficients of

x^{2 n}

on both sides.

Animated Solution for Mathematics - Binomial Theorem: 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

Given Multinomial Expansion

The Substitution Trick: $x \to - \frac{1}{x}$

Simplifying the Substituted Form

Rearranging the Identity

Multiplying the Two Expansions

Simplifying the LHS Product

Extracting Coefficient from RHS

Extracting Coefficient from LHS

Final Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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

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

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 $C_{r} =^{n} C_{r}$ .

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

Let $[x]$ denote greatest integer less than or equal to $x$ . If for $n \in N$ , $(1 - x + x^{3})^{n} = \sum_{j = 0}^{3 n} a_{j} x^{j}$ , then $\sum_{j = 0}^{[\frac{3 n}{2}]} a_{2 j} + 4 \sum_{j = 0}^{[\frac{3 n - 1}{2}]} a_{2 j + 1}$ is equal to :

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 $p < q$ .

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 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

Given Multinomial Expansion

The Substitution Trick: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→−x1​

Simplifying the Substituted Form

Rearranging the Identity

Multiplying the Two Expansions

Simplifying the LHS Product

Extracting Coefficient from RHS

Extracting Coefficient from LHS

Final Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

.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; } 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 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 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; } 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 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.

The Substitution Trick: $x \to - \frac{1}{x}$

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 $C_{r} =^{n} C_{r}$ .

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 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 $p < q$ .

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 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 $m \leq n$ .