Yes, for natural numbers , repeated addition holds directly without needing continuity.

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Sequences and Series: JEE Main Mathematics Question with Solution

Suppose

f

is a function satisfying

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

for all

x, y \in \mathbb{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

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Step-by-step solutionView

Correct answer

The value of $m$ is equal to $10$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

$f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{N}$ , $f(1) = \frac{1}{5}$ , and $\sum_{n=1}^m \frac{f(n)}{n(n+1)(n+2)} = \frac{1}{12}$ .

02goal

Determine the value of the positive integer $m$ .

03approach

Use the Cauchy additive functional equation on $\mathbb{N}$ to deduce $f(n) = n f(1) = \frac{n}{5}$ . Substitute this into the summand to simplify it into $\frac{1}{5(n+1)(n+2)}$ , then use partial fractions to evaluate the telescoping sum.

04execute

Since $f(x+y) = f(x) + f(y)$ for $x, y \in \mathbb{N}$ , by induction we have $f(n) = n f(1) = \frac{n}{5}$ .

05execute

Substitute $f(n) = \frac{n}{5}$ into the general term: $\frac{f(n)}{n(n+1)(n+2)} = \frac{\frac{n}{5}}{n(n+1)(n+2)} = \frac{1}{5(n+1)(n+2)} = \frac{1}{5}\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$

06execute

Evaluate the telescoping sum: $\sum_{n=1}^m \frac{f(n)}{n(n+1)(n+2)} = \frac{1}{5} \sum_{n=1}^m \left(\frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{1}{5} \left( \frac{1}{2} - \frac{1}{m+2} \right)$ Equate this to $\frac{1}{12}$ : $\frac{1}{5} \left( \frac{1}{2} - \frac{1}{m+2} \right) = \frac{1}{12} \implies \frac{1}{2} - \frac{1}{m+2} = \frac{5}{12} \implies \frac{1}{m+2} = \frac{1}{12} \implies m + 2 = 12 \implies m = 10$

✓verify

For $m = 10$ , $\frac{1}{5}\left(\frac{1}{2} - \frac{1}{12}\right) = \frac{1}{5} \cdot \frac{5}{12} = \frac{1}{12}$ , which matches the right-hand side.

Hints that build this answer step by step

Given $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{N}$ and $f(1) = \frac{1}{5}$ , what is the explicit form of $f(n)$ ?
$f(n) = \frac{n}{5}$
Substituting $f(n) = \frac{n}{5}$ into the general term of the summation, how does the summand simplify?
$\frac{1}{5(n+1)(n+2)}$
What is the partial fraction decomposition of $\frac{1}{(n+1)(n+2)}$ and the resulting evaluated sum from $n=1$ to $m$ ?
$\frac{1}{n+1} - \frac{1}{n+2}$ , which telescopes to $\frac{1}{5}\left(\frac{1}{2} - \frac{1}{m+2}\right)$
Solving $\frac{1}{5}\left(\frac{1}{2} - \frac{1}{m+2}\right) = \frac{1}{12}$ yields which value for $m$ ?
$m = 10$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Does

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

always imply

f(n)=n f(1)

Yes, for natural numbers $\mathbb{N}$ , repeated addition $f(n) = f(1 + 1 + \dots + 1) = f(1) + f(1) + \dots + f(1) = n f(1)$ holds directly without needing continuity.

Answer

The value of $m$ is equal to $10$ .

Why each option works or fails

Step-by-step solution

given: $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{N}$ , $f(1) = \frac{1}{5}$ , and $\sum_{n=1}^m \frac{f(n)}{n(n+1)(n+2)} = \frac{1}{12}$ .
goal: Determine the value of the positive integer $m$ .
approach: Use the Cauchy additive functional equation on $\mathbb{N}$ to deduce $f(n) = n f(1) = \frac{n}{5}$ . Substitute this into the summand to simplify it into $\frac{1}{5(n+1)(n+2)}$ , then use partial fractions to evaluate the telescoping sum.
execute: Since $f(x+y) = f(x) + f(y)$ for $x, y \in \mathbb{N}$ , by induction we have $f(n) = n f(1) = \frac{n}{5}$ .
execute: Substitute $f(n) = \frac{n}{5}$ into the general term: $\frac{f(n)}{n(n+1)(n+2)} = \frac{\frac{n}{5}}{n(n+1)(n+2)} = \frac{1}{5(n+1)(n+2)} = \frac{1}{5}\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$
execute: Evaluate the telescoping sum: $\sum_{n=1}^m \frac{f(n)}{n(n+1)(n+2)} = \frac{1}{5} \sum_{n=1}^m \left(\frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{1}{5} \left( \frac{1}{2} - \frac{1}{m+2} \right)$ Equate this to $\frac{1}{12}$ : $\frac{1}{5} \left( \frac{1}{2} - \frac{1}{m+2} \right) = \frac{1}{12} \implies \frac{1}{2} - \frac{1}{m+2} = \frac{5}{12} \implies \frac{1}{m+2} = \frac{1}{12} \implies m + 2 = 12 \implies m = 10$
verify: For $m = 10$ , $\frac{1}{5}\left(\frac{1}{2} - \frac{1}{12}\right) = \frac{1}{5} \cdot \frac{5}{12} = \frac{1}{12}$ , which matches the right-hand side.

Sequences and Series: JEE Main Mathematics Question with Solution

What feels right?

The value of $m$ is equal to $10$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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