JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

The sum $\frac{1 ^{3}}{1} + \frac{1 ^{3} + 2 ^{3}}{1 + 3} + \frac{1 ^{3} + 2 ^{3} + 3 ^{3}}{1 + 3 + 5} + \dots$ up to $8$ terms, is :

Solution

The numerator of the $r$ -th term is $(\frac{r ( r + 1 )}{2})^{2}$ and the denominator is $r^{2}$ (sum of first $r$ odd numbers).

Hence $T_{r} = \frac{( r + 1 ) ^{2}}{4} = \frac{r ^{2} + 2 r + 1}{4}$ .

$S_{n} = \frac{1}{4} [\frac{n ( n + 1 ) ( 2 n + 1 )}{6} + n (n + 1) + n]$ .

For $n = 8$ : $\frac{8 \cdot 9 \cdot 17}{6} = 204$ , $8 \cdot 9 = 72$ , and $n = 8$ .

$S_{8} = \frac{1}{4} (204 + 72 + 8) = \frac{284}{4} = 71$ .

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