If (1+x-2x^2)^6=1+a_1x+a_2x^2+…+a_12x^12, then the value of a_2+a_4+a_6+…+a_12 is:

D. 31. Step 1: Let f(x)=(1+x-2x^2)^6. Then f(1)=0^6=0 and f(-1)=(1-1-2)^6=(-2)^6=64. Step 2: f(1)+f(-1)/2=a_0+a_2+a_4+…+a_12=0+64/2=32. Step 3: The constant term is a_0=f(0)=1, so a_2+a_4+…+a_12=32-1=31. Check: The odd coefficients add to f(1)-f(-1)/2=-32, and a_0+(even)+(odd)=1+31-32=0=f(1). Trap: (C) 32 still contains the constant term a_0=1, which the question leaves out (the source key gave 32).

Question 8 of 13 · Binomial Theorem

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If $(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + \dots + a_{12} x^{12}$ , then the value of $a_{2} + a_{4} + a_{6} + \dots + a_{12}$ is:

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Answer: Option D — 31

Detailed Solution

Let $f (x) = (1 + x - 2 x^{2})^{6}$ . Then $f (1) = 0^{6} = 0$ and $f (- 1) = (1 - 1 - 2)^{6} = (- 2)^{6} = 64$ .

$\frac{f ( 1 ) + f ( - 1 )}{2} = a_{0} + a_{2} + a_{4} + \dots + a_{12} = \frac{0 + 64}{2} = 32$ .

The constant term is $a_{0} = f (0) = 1$ , so $a_{2} + a_{4} + \dots + a_{12} = 32 - 1 = 31$ .

✓ Check

The odd coefficients add to

\frac{f ( 1 ) - f ( - 1 )}{2} = - 32

, and

a_{0} + (e v e n) + (o dd) = 1 + 31 - 32 = 0 = f (1)

⚠ Common trap

a_{0} = 1

, which the question leaves out (the source key gave 32).

Quick tip

$a_{2} + a_{4} + \dots + a_{12} = \frac{f ( 1 ) + f ( - 1 )}{2} - f (0) = \frac{0 + 64}{2} - 1 = 31$

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Binomial Theorem — Preparation

This question tests your understanding of Binomial Theorem, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

The binomial theorem gives the expansion of (a + b)^n for a positive integer n as the sum of terms nCr a^(n-r) b^r. This topic covers the general term T(r+1) = nCr a^(n-r) b^r, finding a specific term or the coefficient of a power of x, the term independent of x, the middle term(s), and properties of binomial coefficients. Almost every question is solved using the general term. Write T(r+1), simplify the power of x, and set it equal to the power you need (or to zero for the term independent of x) to find r. Remember that the expansion has n + 1 terms, so there is one middle term when n is even and two when n is odd. The sum of all binomial coefficients is 2^n. Practise the binomial theorem questions below, each with a step-by-step solution.

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If $(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + \dots + a_{12} x^{12}$ , then the value of $a_{2} + a_{4} + a_{6} + \dots + a_{12}$ is:

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If $(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + \dots + a_{12} x^{12}$ , then the value of $a_{2} + a_{4} + a_{6} + \dots + a_{12}$ is: