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JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Sequences and Series

Suppose f is a function satisfying

f (x + y) = f (x) + f (y)

for all

x, y \in N

and

f (1) = \frac{1}{5}

. If

\sum_{n = 1}^{m} \frac{f ( n )}{n ( n + 1 ) ( n + 2 )} = \frac{1}{12}

, then

m

is equal to ______.

Answer: 10

Solution

The relation

f (x + y) = f (x) + f (y)

on the naturals with

f (1) = \frac{1}{5}

forces

f (n) = n f (1) = \frac{n}{5}

. Hence

\sum_{n = 1}^{m} \frac{f ( n )}{n ( n + 1 ) ( n + 2 )} = \frac{1}{5} \sum_{n = 1}^{m} \frac{1}{( n + 1 ) ( n + 2 )} = \frac{1}{5} \sum_{n = 1}^{m} (\frac{1}{n + 1} - \frac{1}{n + 2}) = \frac{1}{5} (\frac{1}{2} - \frac{1}{m + 2}) = \frac{m}{10 ( m + 2 )}

. Setting

\frac{m}{10 ( m + 2 )} = \frac{1}{12}

gives

12 m = 10 m + 20

, so

m = 10

.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer

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