JEE Mathematics 2023 Question with Solution

Given: The series has general term

$$T_r = \frac{r}{1+r^2+r^4}$$

and we need the sum of the first $10$ terms.

Find:

$$\sum_{r=1}^{10} T_r$$

From the extracted solution,

$$T_r = \frac{2(r^4+r^2+1)}{(r^2+r+1)-(r^2-r+1)}$$

which simplifies to

$$T_r = \frac{1}{2}\left[\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right]$$

Now write the first few terms:

$$T_1 = \frac{1}{2}\left[\frac{1}{1}-\frac{1}{3}\right]$$ $$T_2 = \frac{1}{2}\left[\frac{1}{3}-\frac{1}{7}\right]$$ $$T_3 = \frac{1}{2}\left[\frac{1}{7}-\frac{1}{13}\right]$$

(\dots)

$$T_{10} = \frac{1}{2}\left[\frac{1}{91}-\frac{1}{111}\right]$$

So the sum telescopes:

$$\sum_{r=1}^{10} T_r = \frac{1}{2}\left[1-\frac{1}{111}\right]$$

Hence,

$$\sum_{r=1}^{10} T_r = \frac{1}{2}\cdot\frac{110}{111} = \frac{55}{111}$$

Therefore, the value of the sum is $\frac{55}{111}$.

However, the solution explicitly states The Correct Option is C, while the computed value $\frac{55}{111}$ matches option B in the listed options. Following the solution as primary source, the recorded answer is C.

Given:

$$T_r = \frac{r}{1+r^2+r^4}$$

Find: the sum up to $10$ terms.

Factor the denominator as

$$1+r^2+r^4 = (r^2-r+1)(r^2+r+1)$$

Then

$$T_r = \frac{r}{(r^2-r+1)(r^2+r+1)}$$

Using partial fractions,

$$T_r = \frac{1}{2}\left[\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right]$$

Now observe the pattern:

$$r^2+r+1 = (r+1)^2-(r+1)+1$$

So consecutive terms cancel in the sum.

Thus,

$$\begin{aligned} \sum_{r=1}^{10} T_r &= \frac{1}{2}\sum_{r=1}^{10}\left[\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right] \\ &= \frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\cdots+\left(\frac{1}{91}-\frac{1}{111}\right)\right] \\ &= \frac{1}{2}\left(1-\frac{1}{111}\right) \\ &= \frac{55}{111} \end{aligned}$$

Therefore, the sum equals $\frac{55}{111}$. This matches option B, although the solution labels the correct option as C.