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NVAMediumJEE 2026Sum of Series

JEE Mathematics 2026 Question with Solution

If ∑r=125(rr4+r2+1)=pq,\sum_{r=1}^{25}\left(\frac{r}{r^4+r^2+1}\right)=\frac{p}{q}, where pp and qq are positive integers such that gcd⁡(p,q)=1,\gcd(p,q)=1, then p+qp+q is equal to _____.

Answer

Correct answer:976

Step-by-step solution

Standard Method

Given:

∑r=125rr4+r2+1=pq\sum_{r=1}^{25}\frac{r}{r^4+r^2+1}=\frac{p}{q}

where pp and qq are positive integers with gcd⁡(p,q)=1\gcd(p,q)=1.

Find: p+qp+q.

Step 1: Simplify the general term

r4+r2+1=(r2+r+1)(r2−r+1)r^4+r^2+1=(r^2+r+1)(r^2-r+1)

Now decompose:

rr4+r2+1=12(1r2−r+1−1r2+r+1)\frac{r}{r^4+r^2+1} =\frac{1}{2}\left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right)

Step 2: Use telescoping nature

∑r=125rr4+r2+1=12∑r=125(1r2−r+1−1r2+r+1)\sum_{r=1}^{25}\frac{r}{r^4+r^2+1} =\frac12\sum_{r=1}^{25} \left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right)

Write initial and final terms explicitly:

=12(11−13+13−17+⋯+1601−1651)=\frac12\left( \frac{1}{1}-\frac{1}{3} +\frac{1}{3}-\frac{1}{7} +\cdots +\frac{1}{601}-\frac{1}{651} \right)

All intermediate terms cancel.

=12(1−1651)=12⋅650651=325651=\frac12\left(1-\frac{1}{651}\right) =\frac12\cdot\frac{650}{651} =\frac{325}{651}

Step 3: Compute p+qp+q

p=325, q=651, gcd⁡(325,651)=1p=325,\ q=651,\ \gcd(325,651)=1 p+q=976p+q=976

Therefore, the required value is 976976.

Telescoping Recognition

Given:

∑r=125rr4+r2+1\sum_{r=1}^{25}\frac{r}{r^4+r^2+1}

Find: the value of p+qp+q when the sum is written as pq\frac{p}{q}.

The key observation is that

r4+r2+1=(r2−r+1)(r2+r+1)r^4+r^2+1=(r^2-r+1)(r^2+r+1)

and the numerator rr suggests forming a difference of reciprocals:

rr4+r2+1=12(1r2−r+1−1r2+r+1)\frac{r}{r^4+r^2+1} =\frac12\left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right)

This works because consecutive terms produce cancellation in the series. Hence,

∑r=125rr4+r2+1=12(1−1651)=325651\sum_{r=1}^{25}\frac{r}{r^4+r^2+1} =\frac12\left(1-\frac{1}{651}\right) =\frac{325}{651}

So p=325p=325 and q=651q=651, giving

p+q=976p+q=976

Therefore, the required value is 976976.

Common mistakes

  • A common mistake is to try direct summation term by term without factoring r4+r2+1r^4+r^2+1. This hides the telescoping structure. Instead, first factor the denominator as $$$(r^2+r+1)(r^2-r+1)$$ and then decompose the term.

  • Another mistake is using an incorrect partial fraction form, such as missing the factor 12\frac12. That gives the wrong surviving terms after cancellation. Verify the identity carefully before summing.

  • Some students stop after obtaining 325651\frac{325}{651} and report that as the final answer. This is wrong because the question asks for p+qp+q, not the value of the sum itself. After identifying p=325p=325 and q=651q=651, compute 325+651325+651.

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