If the sum of the first 10 terms of the series , is , gcd (m, n) = 1, then m + n is equal to: [2026]

Question

If the sum of the first 10 terms of the series $\frac{1}{1 + 1^{4} \times 4} + \frac{2}{1 + 2^{4} \times 4} + \frac{3}{1 + 3^{4} \times 4} + \frac{4}{1 + 4^{4} \times 4} + . . .$ , is $\frac{m}{n}$ , gcd (m, n) = 1, then m + n is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

Given series can be written in general form as

$\sum_{r = 1}^{10} (\frac{r}{1 + 4 r^{4}})$

Now, $1 + 4 r^{4} = (2 r^{2} - 2 r + 1) (2 r^{2} + 2 r + 1)$

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

$∴ \sum_{r = 1}^{r = 10} (\frac{r}{1 + 4 r^{4}})$

$= \sum_{r = 1}^{r = 10} \frac{1}{4} (\frac{1}{2 r^{2} - 2 r + r} - \frac{1}{2 r^{2} + 2 r + r})$

$= \frac{1}{4} (1 - \frac{1}{221}) = \frac{55}{221}$

$∴ \frac{m}{n} = \frac{55}{221} \Rightarrow m = 55 and n = 221$

$∴ m + n = 276$

Solution

Q.
If the sum of the first 10 terms of the series $\frac{1}{1 + 1^{4} \times 4} + \frac{2}{1 + 2^{4} \times 4} + \frac{3}{1 + 3^{4} \times 4} + \frac{4}{1 + 4^{4} \times 4} + . . .$ , is $\frac{m}{n}$ , gcd (m, n) = 1, then m + n is equal to: [2026]

1 256

2 264

3 276

4 284

Want Us to Email you About Special Offers & Updates?