NIMCET 2021 — Mathematics PYQ

We know the identity $1+x^3=(1+x)(1-x+x^2)$. Taking the natural logarithm on both sides:

Therefore:

2. Expand using Power Series

Using the standard expansion $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$, we get:

3. Identify the Coefficients

Subtracting the two series:

We are interested in the coefficients where the power of $x$ is a multiple of 3 (i.e., $a_3,a_6,a_9,\dots$).

• For $x^3$: $a_3=1-\frac{1}{3}=\frac{2}{3}$

• For $x^6$: $a_6=-\frac{1}{2}-(-\frac{1}{6})=-\frac{1}{2}+\frac{1}{6}=-\frac{2}{6}=-\frac{1}{3}=\frac{2}{3}\cdot (-\frac{1}{2})$

• For $x^9$: $a_9=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}=\frac{2}{3}\cdot (\frac{1}{3})$

4. Sum the Series

The required sum is $S=a_3+a_6+a_9+\dots$:

Wait, let's look at the general form of $a_{3k}$:

Since $(-1{)}^{3k-1}$ is the same as $(-1{)}^{k-1}$ only if we track parity carefully, let's simplify:

Now, sum from $k=1$ to $\infty$:

We know that ${\sum }_{k=1}^{\infty }\frac{(-1{)}^{k-1}}{k}=\log 2$.

Result:

Correct Option: (B)