JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q6

MathematicsSingle correct+4 / -1Question 6 of 75

The sum $1+\dfrac12\left(1^2+2^2\right)+\dfrac13\left(1^2+2^2+3^2\right)+\dots$ upto 10 terms is equal to :

130

155

\dfrac{315}2

\dfrac{325}2

Answer and worked solution

Official answer: C

General term.
$T_n=\frac1n\cdot\frac{n(n+1)(2n+1)}6=\frac{2n^2+3n+1}6.$

Sum to 10 terms.
$\frac{2(385)+3(55)+10}6=\frac{945}6=\frac{315}2.$

(C)

Topics:Special series sums

JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 6