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JEE Main 2021
Binomial Theorem
Binomial Theorem
Easy

Question

If A denotes the sum of all the coefficients in the expansion of (13x+10x2)n\left(1-3 x+10 x^2\right)^{\mathrm{n}} and B denotes the sum of all the coefficients in the expansion of (1+x2)n\left(1+x^2\right)^n, then :

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Solution

Key Concept: Sum of Coefficients in a Polynomial Expansion

The sum of all coefficients in the expansion of any polynomial function P(x)P(x) can be easily found by substituting x=1x=1 into the polynomial. This is because when x=1x=1, any term ckxkc_k x^k simply becomes ck(1)k=ckc_k (1)^k = c_k. Thus, evaluating P(1)P(1) yields the sum of all its coefficients.

Step-by-Step Derivation of A

  1. Identify the Polynomial: We are given that A denotes the sum of all coefficients in the expansion of (13x+10x2)n\left(1-3 x+10 x^2\right)^n. Let P(x)=(13x+10x2)nP(x) = \left(1-3 x+10 x^2\right)^n.

  2. Apply the Key Concept: To find the sum of coefficients, which is A, we substitute x=1x=1 into the polynomial P(x)P(x). A=(13(1)+10(1)2)nA = \left(1-3(1)+10(1)^2\right)^n

  3. Simplify the Expression: Perform the arithmetic operations inside the parentheses. A=(13+10)nA = (1 - 3 + 10)^n A=(8)nA = (8)^n So, we have A=8nA = 8^n.

Step-by-Step Derivation of B

  1. Identify the Polynomial: Similarly, B denotes the sum of all coefficients in the expansion of (1+x2)n\left(1+x^2\right)^n. Let Q(x)=(1+x2)nQ(x) = \left(1+x^2\right)^n.

  2. Apply the Key Concept: To find the sum of coefficients, which is B, we substitute x=1x=1 into the polynomial Q(x)Q(x). B=(1+(1)2)nB = \left(1+(1)^2\right)^n

  3. Simplify the Expression: Perform the arithmetic operations inside the parentheses. B=(1+1)nB = (1+1)^n B=(2)nB = (2)^n So, we have B=2nB = 2^n.

Establishing the Relationship between A and B

Now that we have expressions for A and B in terms of nn: A=8nA = 8^n B=2nB = 2^n We need to find how A and B are related. Let's express 8n8^n using the base 22. We know that 8=238 = 2^3.

  1. Substitute and Rewrite A: Replace 88 with 232^3 in the expression for A. A=(23)nA = (2^3)^n

  2. Apply Exponent Rule: Using the exponent rule (am)n=amn(a^m)^n = a^{mn}, we multiply the exponents. A=23nA = 2^{3n}

  3. Relate to B: We can rewrite 23n2^{3n} as (2n)3(2^n)^3. A=(2n)3A = (2^n)^3 Since we found that B=2nB = 2^n, we can substitute B into this equation. A=B3A = B^3

Tips for Success and Common Pitfalls

  • Universality: Remember that this method (P(1)P(1)) works for finding the sum of coefficients for any polynomial, not just binomial expansions.
  • Careful Substitution: Double-check your arithmetic when substituting x=1x=1, especially with negative signs.
  • Distinguish from Binomial Theorem: While this problem involves binomial expansions, the core concept used (sum of coefficients) is distinct from finding specific terms or properties of binomial coefficients.
  • Other Sums: If a question asks for the sum of coefficients of terms with even powers or odd powers, a different approach involving P(1)P(1) and P(1)P(-1) would be needed.

Summary and Key Takeaway

By applying the fundamental principle that the sum of coefficients of a polynomial P(x)P(x) is simply P(1)P(1), we efficiently calculated A=8nA=8^n and B=2nB=2^n. Subsequently, recognizing that 88 is 232^3, we were able to establish the relationship A=B3A = B^3. This highlights the importance of understanding basic algebraic properties and exponent rules in simplifying and solving problems.

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