MCQEasyJEE 2024Simple Applications

JEE Mathematics 2024 Question with Solution

If AA denotes the sum of all the coefficients in the expansion of (13x+10x2)n(1 - 3x + 10x^2)^n and BB denotes the sum of all the coefficients in the expansion of (1+x2)n(1 + x^2)^n, then:

  • A

    A=B3A = B^3

  • B

    3A=B3A = B

  • C

    B=A3B = A^3

  • D

    A=3BA = 3B

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

  • AA is the sum of all coefficients in the expansion of (13x+10x2)n(1 - 3x + 10x^2)^n.
  • BB is the sum of all coefficients in the expansion of (1+x2)n(1 + x^2)^n.

Find: The relation between AA and BB.

The sum of coefficients of a polynomial is obtained by substituting x=1x = 1.

So,

A=(131+1012)n=(13+10)n=8nA = (1 - 3 \cdot 1 + 10 \cdot 1^2)^n = (1 - 3 + 10)^n = 8^n

Also,

B=(1+12)n=2nB = (1 + 1^2)^n = 2^n

Now,

A=8n=(23)n=(2n)3=B3A = 8^n = (2^3)^n = (2^n)^3 = B^3

Therefore, the correct option is A, that is, A=B3A = B^3.

Step-by-step Evaluation

Given: The expressions are (13x+10x2)n(1 - 3x + 10x^2)^n and (1+x2)n(1 + x^2)^n.

Find: Values of AA and BB in terms of nn, and then compare them.

  1. For (13x+10x2)n(1 - 3x + 10x^2)^n, substitute x=1x = 1 to get the sum of coefficients.
A=(131+1012)nA = (1 - 3 \cdot 1 + 10 \cdot 1^2)^n A=(13+10)n=8nA = (1 - 3 + 10)^n = 8^n
  1. For (1+x2)n(1 + x^2)^n, again substitute x=1x = 1.
B=(1+12)n=2nB = (1 + 1^2)^n = 2^n
  1. Compare the two values.
8n=(23)n=(2n)38^n = (2^3)^n = (2^n)^3

Hence,

A=B3A = B^3

Therefore, the required relation is A=B3A = B^3.

Common mistakes

  • A common mistake is to try to expand the binomials or trinomial fully. That is unnecessary because the sum of coefficients is found by putting x=1x = 1. Use substitution instead of expansion.

  • Some students substitute x=0x = 0 instead of x=1x = 1. Substituting x=0x = 0 gives only the constant term, not the sum of all coefficients. Always use x=1x = 1 for coefficient sum.

  • Another mistake is comparing A=8nA = 8^n and B=2nB = 2^n incorrectly. Since 8=238 = 2^3, we get 8n=(2n)38^n = (2^n)^3, so A=B3A = B^3, not B=A3B = A^3.

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