JEE Main 2022 · 27 Jul · Shift 2 · Mathematics Q24

← JEE Main 2022 · 27 Jul · Shift 2

MathematicsInteger+4 / -1Question 24 of 90

$\dfrac{2^3-1^3}{1\times7}+\dfrac{4^3-3^3+2^3-1^3}{2\times11}+\dfrac{6^3-5^3+4^3-3^3+2^3-1^3}{3\times15}+\cdots+\dfrac{30^3-29^3+28^3-27^3+\ldots+2^3-1^3}{15\times63}$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 120

Numerator. $\sum_{k=1}^n\left[(2k)^3-(2k-1)^3\right]=n^2(4n+3)$ , and the denominator is $n(4n+3)$ , so the n-th term is n.

Sum $1+\dots+15=120$ .

Answer: 120

Topics:Special series sums

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JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 24