JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The n-th term of the given series is: T_n = 1 n (1^2 + 2^2 + + n^2) Using the formula for the sum of squares of the first n natural numbers: T_n = 1 n ( n(n+1)(2n+1) 6 ) = (n+1)(2n+1) 6 = 2n^2 + 3n + 1 6 The sum of the first 10 terms is S_ 10 = _ n=1 ^ 10 T_n S_ 10 = 1 6 ( 2 _ n=1 ^ 10 n^2 + 3 _ n=1 ^ 10 n + _ n=1 ^ 10 1 ) Using the standard summation formulas: _ n=1 ^ 10 n^2 = 10 11 21 6 = 385 _ n=1 ^ 10 n = 10 11 2 = 55 _ n=1 ^ 10 1 = 10 Substituting these values into the sum expression: S_ 10 = 1 6 (2 385 + 3 55 + 10) S_ 10 = 1 6 (770 + 165 + 10) S_ 10 = 945 6 = 315 2 Answer: 315 2