Logo
Jeeify
MCQEasyJEE 2023Sum of Series

JEE Mathematics 2023 Question with Solution

Let f(x)f(x) be a function such that f(x+y)=f(x)f(y)f(x + y) = f(x)f(y) for all x,y∈Nx, y \in \mathbb{N}. If f(1)=3f(1) = 3 and ∑k=1nf(k)=3279\sum_{k=1}^{n} f(k) = 3279, then the value of nn is:

  • A

    66

  • B

    88

  • C

    77

  • D

    99

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) for all x,y∈Nx,y\in\mathbb{N}, f(1)=3f(1)=3, and

∑k=1nf(k)=3279\sum_{k=1}^{n} f(k)=3279

Find: the value of nn.

Using the functional equation,

f(2)=f(1+1)=f(1)f(1)=32f(2)=f(1+1)=f(1)f(1)=3^2 f(3)=f(1)f(2)=33f(3)=f(1)f(2)=3^3 f(4)=34f(4)=3^4

Hence,

f(k)=3kf(k)=3^k

for all natural numbers kk.

Therefore,

∑k=1n3k=3279\sum_{k=1}^{n} 3^k=3279

This is a geometric progression with first term 33 and common ratio 33, so

Sn=3(3n−1)3−1S_n=\frac{3(3^n-1)}{3-1}

Thus,

3(3n−1)2=3279\frac{3(3^n-1)}{2}=3279 3n−1=2⋅32793=21863^n-1=\frac{2\cdot 3279}{3}=2186 3n=21873^n=2187

Since

2187=372187=3^7

we get

n=7n=7

Therefore, the correct option is C.

Using the summation formula

Given: f(x+y)=f(x)f(y)f(x+y)=f(x)f(y), f(1)=3f(1)=3, and

∑k=1nf(k)=3279\sum_{k=1}^{n} f(k)=3279

Find: nn.

From the relation,

f(2)=f(1+1)=f(1)f(1)=32=9f(2)=f(1+1)=f(1)f(1)=3^2=9 f(3)=f(2+1)=f(2)f(1)=33=27f(3)=f(2+1)=f(2)f(1)=3^3=27 f(4)=f(3+1)=f(3)f(1)=34=81f(4)=f(3+1)=f(3)f(1)=3^4=81

So the pattern gives

f(k)=3kf(k)=3^k

Now substitute into the sum:

∑k=1n3k=3279\sum_{k=1}^{n} 3^k=3279

Using the GP sum formula,

Sn=a(rn−1)r−1S_n=\frac{a(r^n-1)}{r-1}

with a=3a=3 and r=3r=3,

3279=3(3n−1)23279=\frac{3(3^n-1)}{2} 6558=3(3n−1)6558=3(3^n-1) 3n−1=21863^n-1=2186 3n=21873^n=2187

Since 37=21873^7=2187, we obtain n=7n=7. Therefore, the correct option is C.

Common mistakes

  • Assuming f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) instead of f(x)f(y)f(x)f(y) is incorrect because this functional equation generates an exponential form. Use repeated application with f(1)=3f(1)=3 to get f(k)=3kf(k)=3^k.

  • Using the geometric series formula starting from k=0k=0 is wrong here because the sum begins at k=1k=1. The first term is 33, not 11.

  • Forgetting to divide by 33 after multiplying both sides by 22 leads to an incorrect value of 3n−13^n-1. From 3(3n−1)2=3279\frac{3(3^n-1)}{2}=3279, first write 6558=3(3n−1)6558=3(3^n-1) and then divide by 33.

Practice more Sum of Series questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Sum of Series

Related questions