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Cauchy Functional Equation and Telescoping Sum Decomposition

Q: Does f(x+y)=f(x)+f(y) always imply f(n)=n f(1)?

Yes, for natural numbers ℕ, repeated addition f(n) = f(1 + 1 + … + 1) = f(1) + f(1) + … + f(1) = n f(1) holds directly without needing continuity.

Suppose f is a function satisfying f(x+y)=f(x)+f(y) for all x, y ∈ ℕ and f(1)=1/5. If sum from n=1 to m of (f(n))/(n(n+1)(n+2))=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 $x, y \in N$ , $f (1) = \frac{1}{5}$ , 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 $N$ 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 $x, y \in N$ , by induction we have $f (n) = n f (1) = \frac{n}{5}$ .

05Execute

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

06Execute

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

✓Verify

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

Hints that build this answer step by step

Given $f (x + y) = f (x) + f (y)$ for all $x, y \in N$ 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 $\frac{1}{5} (\frac{1}{2} - \frac{1}{m + 2})$
Solving $\frac{1}{5} (\frac{1}{2} - \frac{1}{m + 2}) = \frac{1}{12}$ yields which value for $m$ ?
$m = 10$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $N$ , 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 $x, y \in N$ , $f (1) = \frac{1}{5}$ , 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 $N$ 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 $x, y \in N$ , by induction we have $f (n) = n f (1) = \frac{n}{5}$ .
Execute: Substitute $f (n) = \frac{n}{5}$ into the general term: $\frac{f (n)}{n (n + 1) (n + 2)} = \frac{\frac{n}{5}}{n (n + 1) (n + 2)} = \frac{1}{5 (n + 1) (n + 2)} = \frac{1}{5} (\frac{1}{n + 1} - \frac{1}{n + 2})$
Execute: Evaluate the telescoping sum: $\sum_{n = 1}^{m} \frac{f (n)}{n (n + 1) (n + 2)} = \frac{1}{5} \sum_{n = 1}^{m} (\frac{1}{n + 1} - \frac{1}{n + 2}) = \frac{1}{5} (\frac{1}{2} - \frac{1}{m + 2})$ Equate this to $\frac{1}{12}$ : $\frac{1}{5} (\frac{1}{2} - \frac{1}{m + 2}) = \frac{1}{12} ⟹ \frac{1}{2} - \frac{1}{m + 2} = \frac{5}{12} ⟹ \frac{1}{m + 2} = \frac{1}{12} ⟹ m + 2 = 12 ⟹ m = 10$
Verify: For $m = 10$ , $\frac{1}{5} (\frac{1}{2} - \frac{1}{12}) = \frac{1}{5} \cdot \frac{5}{12} = \frac{1}{12}$ , which matches the right-hand side.

Cauchy Functional Equation and Telescoping Sum Decomposition

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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The value of mm is equal to 1010.

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02Goal

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04Execute

05Execute

06Execute

✓Verify

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