Sum to 10 Terms of the Series r/(1+r²+r⁴), JEE Main 2023

JEE Main 2023 — 1 February, Morning Shift. Previous Year Question.

Problem

The sum to 10 terms of the series

$$\frac{1}{1+1^2+1^4} + \frac{2}{1+2^2+2^4} + \frac{3}{1+3^2+3^4} + \cdots$$

is:

Key insight. The denominator $1+r^2+r^4$ isn’t just “some quartic” — it’s a disguised difference of squares, $(1+r^2)^2 – r^2$. Once it’s factored that way, the fraction splits into a telescoping pair, and nearly every term in the sum cancels.

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Approach

Write the general term $T_r = \dfrac{r}{1+r^2+r^4}$. Since the denominator doesn’t factor as an obvious product at first glance, add and subtract $r^2$ to reshape it into a difference of squares. That factorisation is the key that unlocks a partial-fraction split designed to telescope — the whole point being that most of the 10 terms then cancel against each other, leaving only the first and last.

Solution

Step 1 — Factor the denominator as a difference of squares

$$1 + r^2 + r^4 = (1+r^2)^2 – r^2 = \big(1+r^2-r\big)\big(1+r^2+r\big)$$

Step 2 — Split into partial fractions

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

This is checked by recombining the right-hand side over a common denominator: the numerator comes out to $(1+r^2+r) – (1+r^2-r) = 2r$, matching $2T_r$.

Step 3 — Spot the telescoping pattern

Let $f(r) = 1+r^2-r$. Then:

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

So $f(r+1)$ is exactly the second denominator term above, which means:

$$T_r = \frac{1}{2}\left[\frac{1}{f(r)} – \frac{1}{f(r+1)}\right]$$

Each term’s “second half” cancels the next term’s “first half” when summed.

Step 4 — Sum the telescoping series to 10 terms

$$\sum_{r=1}^{10} T_r = \frac{1}{2}\left[\frac{1}{f(1)} – \frac{1}{f(11)}\right]$$

With $f(1) = 1+1-1 = 1$ and $f(11) = 1+121-11 = 111$:

$$\sum_{r=1}^{10} T_r = \frac{1}{2}\left[\frac{1}{1} – \frac{1}{111}\right] = \frac{1}{2} \cdot \frac{110}{111} = \frac{55}{111}$$

Answer

Common mistakes

• Not recognising $1+r^2+r^4$ as a difference of squares. Without the “add and subtract $r^2$” trick, the denominator looks unfactorable, and the telescoping structure never appears.
• Losing track of which term is $f(r)$ and which is $f(r+1)$. Since $f(r+1)$ replaces $r$ with $r+1$ in $f(r) = 1+r^2-r$, it’s worth explicitly verifying $f(r+1) = 1+r^2+r$ (as done in Step 3) rather than assuming the pattern.

Practise next

• Find the sum to 15 terms of the same series $\displaystyle\sum \frac{r}{1+r^2+r^4}$, using the same telescoping identity.