If the sum of the first 20 terms of the series , where m and n are coprime, then m + n is equal to : [2025]

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Solution

Q.
If the sum of the first 20 terms of the series $\frac{4 \cdot 1}{4 + 3 \cdot 1^{2} + 1^{4}} + \frac{4 \cdot 2}{4 + 3 \cdot 2^{2} + 2^{4}} + \frac{4 \cdot 3}{4 + 3 \cdot 3^{2} + 3^{4}} + \frac{4 \cdot 4}{4 + 3 \cdot 4^{2} + 4^{4}} + . . . is \frac{m}{n}$ , where m and n are coprime, then m + n is equal to : [2025]

1 423

2 421

3 420

4 422

Ans.
(2)

$\frac{4 \cdot 1}{4 + 3 \cdot 1^{2} + 1^{4}} + \frac{4 \cdot 2}{4 + 3 \cdot 2^{2} + 2^{4}} + . . . .$ upto 20 terms

$= \sum_{r = 1}^{20} \frac{4 r}{4 + 3 r^{2} + r^{4}}$

$= \sum_{r = 1}^{20} \frac{4 r}{(r^{2} + r + 2) (r^{2} - r + 2)}$

$= 2 \sum_{r = 1}^{20} \frac{1}{r^{2} - r + 2} - \frac{1}{r^{2} + r + 2}$

$= 2 [(\frac{1}{2} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{8}) + (\frac{1}{8} - \frac{1}{14}) + . . . + (\frac{1}{382} - \frac{1}{422})]$

$= 2 [\frac{1}{2} - \frac{1}{422}] = \frac{420}{422} = \frac{210}{211} = \frac{m}{n}$ , m and n are co-prime.

So, m + n = 210 + 211 = 421.

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