If the sum of the first 10 terms of the series , where gcd (m, n) = 1, then m + n is equal to __________. [2025]

Question

If the sum of the first 10 terms of the series $\frac{4 \cdot 1}{1 + 4 \cdot 1^{4}} + \frac{4 \cdot 2}{1 + 4 \cdot 2^{4}} + \frac{4 \cdot 3}{1 + 4 \cdot 3^{4}} + . . . is \frac{m}{n}$ , where gcd (m, n) = 1, then m + n is equal to __________. [2025]

Smart Achievers · Accepted Answer

(441)

The given series is

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

$∴$ General term of series is $\frac{4 k}{1 + 4 k^{4}}$

Now, $\frac{4 k}{1 + 4 k^{4}} = \frac{4 k}{(2 k^{2} - 2 k + 1) (2 k^{2} + 2 k + 1)}$

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

$= \frac{1}{2 k^{2} - 2 k + 1} - \frac{1}{2 k^{2} + 2 k + 1}$

Now, $S_{10} = \sum_{k = 1}^{10} \frac{4 k}{1 + 4 k^{4}}$

$= \sum_{k = 1}^{10} [\frac{1}{2 k^{2} - 2 k + 1} - \frac{1}{2 k^{2} + 2 k + 1}]$

$\sum_{k = 1}^{10} [\frac{1}{2 k^{2} - 2 k + 1} - \frac{1}{2 {(k + 1)}^{2} - 2 (k + 1) + 1}]$

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

$= 1 - \frac{1}{242 - 22 + 1} = 1 - \frac{1}{221} = \frac{220}{221}$

$∴ \frac{m}{n} = \frac{220}{221} \Rightarrow m + n = 220 + 221 = 441$ .

Solution

Q.
If the sum of the first 10 terms of the series $\frac{4 \cdot 1}{1 + 4 \cdot 1^{4}} + \frac{4 \cdot 2}{1 + 4 \cdot 2^{4}} + \frac{4 \cdot 3}{1 + 4 \cdot 3^{4}} + . . . is \frac{m}{n}$ , where gcd (m, n) = 1, then m + n is equal to __________. [2025]

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Solution

Q. If the sum of the first 10 terms of the series 4·11+4·14+4·21+4·24+4·31+4·34+... is mn, where gcd (m, n) = 1, then m + n is equal to __________. [2025]

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Q.
If the sum of the first 10 terms of the series $\frac{4 \cdot 1}{1 + 4 \cdot 1^{4}} + \frac{4 \cdot 2}{1 + 4 \cdot 2^{4}} + \frac{4 \cdot 3}{1 + 4 \cdot 3^{4}} + . . . is \frac{m}{n}$ , where gcd (m, n) = 1, then m + n is equal to __________. [2025]