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NVAMediumJEE 2026Definite Integrals

JEE Mathematics 2026 Question with Solution

Let [⋅][ \cdot ] denote the greatest integer function and f(x)=lim⁡n→∞1n3∑k=1n[k23x]f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^{n} \left[ \frac{k^2}{3^x} \right]. Then 12∑j=1∞f(j)12 \sum_{j=1}^{\infty} f(j) is equal to _____.

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given:

f(x)=lim⁡n→∞1n3∑k=1n[k23x]f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^{n} \left[ \frac{k^2}{3^x} \right]

We need to find

12∑j=1∞f(j)12 \sum_{j=1}^{\infty} f(j)

Find: The numerical value of the given expression.

Using the property of the greatest integer function,

z−1<[z]≤zz - 1 < [z] \le z

put

z=k23xz = \frac{k^2}{3^x}

Then

k23x−1<[k23x]≤k23x\frac{k^2}{3^x} - 1 < \left[ \frac{k^2}{3^x} \right] \le \frac{k^2}{3^x}

Dividing by n3n^3 and summing from k=1k=1 to nn,

1n3∑k=1n(k23x−1)<1n3∑k=1n[k23x]≤1n3∑k=1nk23x\frac{1}{n^3} \sum_{k=1}^{n} \left( \frac{k^2}{3^x} - 1 \right) < \frac{1}{n^3} \sum_{k=1}^{n} \left[ \frac{k^2}{3^x} \right] \le \frac{1}{n^3} \sum_{k=1}^{n} \frac{k^2}{3^x}

Now

1n3∑k=1n1=nn3=1n2→0\frac{1}{n^3} \sum_{k=1}^{n} 1 = \frac{n}{n^3} = \frac{1}{n^2} \to 0

so the greatest integer part does not affect the limit.

Hence,

f(x)=lim⁡n→∞1n3∑k=1nk23x=13xlim⁡n→∞1n∑k=1n(kn)2\begin{aligned} f(x) &= \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^{n} \frac{k^2}{3^x} \\ &= \frac{1}{3^x} \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^2 \end{aligned}

This is a Riemann sum, so

f(x)=13x∫01t2 dtf(x) = \frac{1}{3^x} \int_0^1 t^2 \, dt

Therefore,

f(x)=13x[t33]01=13x⋅13=13x+1\begin{aligned} f(x) &= \frac{1}{3^x} \left[ \frac{t^3}{3} \right]_0^1 \\ &= \frac{1}{3^x} \cdot \frac{1}{3} \\ &= \frac{1}{3^{x+1}} \end{aligned}

Now,

∑j=1∞f(j)=∑j=1∞13j+1=132+133+⋯\sum_{j=1}^{\infty} f(j) = \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} = \frac{1}{3^2} + \frac{1}{3^3} + \cdots

This is a geometric progression with first term

a=19a = \frac{1}{9}

and common ratio

r=13r = \frac{1}{3}

So,

∑j=1∞f(j)=a1−r=1/91−1/3=16\sum_{j=1}^{\infty} f(j) = \frac{a}{1-r} = \frac{1/9}{1-1/3} = \frac{1}{6}

Hence,

12∑j=1∞f(j)=12×16=212 \sum_{j=1}^{\infty} f(j) = 12 \times \frac{1}{6} = 2

Therefore, the required value is 22. The solution's lists correct answer 44, but the extracted working gives 22, and the solution is taken.

Common mistakes

  • Treating [k23x]\left[ \frac{k^2}{3^x} \right] as exactly equal to k23x\frac{k^2}{3^x} without justification. This is incomplete because the floor function introduces an error. Use z−1<[z]≤zz-1<[z]\le z and show that after division by n3n^3 the total error is nn3=1n2→0\frac{n}{n^3}=\frac{1}{n^2} \to 0.

  • Missing the Riemann sum structure. The term 1n3∑k2\frac{1}{n^3}\sum k^2 should be rewritten as 1n∑(kn)2\frac{1}{n}\sum \left(\frac{k}{n}\right)^2, which leads to the integral ∫01t2 dt\int_0^1 t^2 \, dt. Without this step, the limit is harder to evaluate correctly.

  • Starting the geometric series from the wrong index. Since f(j)=13j+1f(j)=\frac{1}{3^{j+1}} and the sum begins at j=1j=1, the first term is 19\frac{1}{9}, not 13\frac{1}{3}. Use the correct first term before applying the GP sum formula.

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