Evaluate 1 2^2 + 2 3^2 + + 100 (101)^2 / (1^2 2^2 + 2^2 3^2 + + 100^2 101):

Evaluate 1 2^2 + 2 3^2 + + 100 (101)^2 / (1^2 2^2 + 2^2 3^2 + + 1… — Solution | JEE Mathematics

MCQMediumJEE 2024Sum of Series

Evaluate $1\times 2^2 + 2\times 3^2 + \ldots + 100\times (101)^2 \, / \, (1^2\times 2^2 + 2^2\times 3^2 + \ldots + 100^2\times 101)$ :

Correct answer:B

Standard Method

Given: Evaluate the ratio

\frac{1\times 2^2 + 2\times 3^2 + \ldots + 100\times 101^2}{1^2\times 2^2 + 2^2\times 3^2 + \ldots + 100^2\times 101}

Find: The correct option.

Let

N = \sum_{k=1}^{100} k(k+1)^2

and

D = \sum_{k=1}^{100} k^2(k+1)

Using summation formulas for the numerator and denominator, compute the ratio. The given solution states that this ratio evaluates to $305$ .

Therefore, the correct option is B.

Expanding only the numerator and not the denominator consistently can distort the ratio. Write both sums in sigma form first, then apply the correct summation formulas to each.
Confusing $k(k+1)^2$ with $k^2(k+1)$ is a structural error because the powers belong to different factors in numerator and denominator. Keep the two series separate before simplification.
Using incorrect formulas for $\sum k^2$ or $\sum k^3$ leads to a wrong final value. Verify the standard summation identities before substituting $n=100$ .

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