The sum 1 + 1 2 (1^2 + 2^2) + 1 3 (1^2 + 2^2 + 3^2) + upto 10 terms is equal to :

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 S₁₀ = _ n=1 ¹⁰ T_n S₁₀ = 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: S₁₀ = 1 6 (2 385 + 3 55 + 10) S₁₀ = 1 6 (7