If log(1-x+x^2)=a_1x+a_2x^2+a_3x^3+…, then a_3+a_6+a_9+… is equal to

B. 2/3log2. Step 1: 1-x+x^2=1+x^31+x, so log(1-x+x^2)=log(1+x^3)-log(1+x). Step 2: The coefficient of x^3k in log(1+x^3) is (-1)^k+1k; in log(1+x) it is (-1)^3k+13k=(-1)^k+13k. Step 3: So a_3k=2/3⋅(-1)^k+1k and a_3+a_6+a_9+…=2/3(1-1/2+1/3-…)=2/3log2. Check: a_3 directly: log(1+x^3) gives +x^3 and -log(1+x) gives -x^33, so a_3=2/3, matching k=1. Trap: (A) log2 comes from using only log(1+x^3) and forgetting to subtract log(1+x).

Question 9 of 15 · Logarithm

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If $l o g (1 - x + x^{2}) = a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots$ , then $a_{3} + a_{6} + a_{9} + \dots$ is equal to

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — $\frac{2}{3} l o g 2$

Detailed Solution

$1 - x + x^{2} = \frac{1 + x ^{3}}{1 + x}$ , so $l o g (1 - x + x^{2}) = l o g (1 + x^{3}) - l o g (1 + x)$ .

The coefficient of $x^{3 k}$ in $l o g (1 + x^{3})$ is $\frac{( - 1 ) ^{k + 1}}{k}$ ; in $l o g (1 + x)$ it is $\frac{( - 1 ) ^{3 k + 1}}{3 k} = \frac{( - 1 ) ^{k + 1}}{3 k}$ .

So $a_{3 k} = \frac{2}{3} \cdot \frac{( - 1 ) ^{k + 1}}{k}$ and $a_{3} + a_{6} + a_{9} + \dots = \frac{2}{3} (1 - \frac{1}{2} + \frac{1}{3} - \dots) = \frac{2}{3} l o g 2$ .

✓ Check

a_{3}

directly:

l o g (1 + x^{3})

gives

+ x^{3}

and

- l o g (1 + x)

gives

- \frac{x ^{3}}{3}

, so

a_{3} = \frac{2}{3}

, matching

k = 1

⚠ Common trap

(A)

l o g 2

comes from using only

l o g (1 + x^{3})

and forgetting to subtract

l o g (1 + x)

Quick tip

$1 - x + x^{2} = \frac{1 + x ^{3}}{1 + x}$ , so $a_{3 k} = \frac{2}{3} \cdot \frac{( - 1 ) ^{k + 1}}{k}$ and the sum is $\frac{2}{3} ln 2$

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Logarithm — Preparation

This question tests your understanding of Logarithm, 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.

A logarithm answers the question: to what power must a base be raised to give a number? If a^x = N, then log base a of N = x. This topic covers the laws of logarithms (product, quotient and power rules), change of base, common and natural logarithms, and solving logarithmic and exponential equations. Most exam questions reduce to applying the three main rules correctly: log(mn) = log m + log n, log(m/n) = log m - log n, and log(m^k) = k log m. Remember that logarithms are only defined for positive numbers and a base that is positive and not equal to 1, which helps you reject invalid answers. Practise the logarithm questions below. Each comes with a step-by-step solution showing which law is applied at each step.

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If $l o g (1 - x + x^{2}) = a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots$ , then $a_{3} + a_{6} + a_{9} + \dots$ is equal to

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