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

Correct answer: C. (315)/(2) 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)

JEE Main 20266 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 6 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2026 · 6 April · Evening Shift · Mathematics · Sequences and Series

The sum 1 + (1)/(2)(1^2 + 2^2) + (1)/(3)(1^2 + 2^2 + 3^2) + upto 10...

The sum

1 + \dfrac{1}{2}(1^2 + 2^2) + \dfrac{1}{3}(1^2 + 2^2 + 3^2) + \ldots

upto 10 terms is equal to :

Options

A $130$
B $155$
C $\dfrac{315}{2}$
D $\dfrac{325}{2}$

Correct answer

C. $\dfrac{315}{2}$

Step-by-step solution

The

n

-th term of the given series is:

T_n = \dfrac{1}{n} (1^2 + 2^2 + \ldots + n^2)

Using the formula for the sum of squares of the first

n

natural numbers:

T_n = \dfrac{1}{n} \left( \dfrac{n(n+1)(2n+1)}{6} \right) = \dfrac{(n+1)(2n+1)}{6} = \dfrac{2n^2 + 3n + 1}{6}

The sum of the first 10 terms is

S_{10} = \sum_{n=1}^{10} T_n

S_{10} = \dfrac{1}{6} \left( 2 \sum_{n=1}^{10} n^2 + 3 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 1 \right)

Using the standard summation formulas:

\sum_{n=1}^{10} n^2 = \dfrac{10 \times 11 \times 21}{6} = 385

\sum_{n=1}^{10} n = \dfrac{10 \times 11}{2} = 55

\sum_{n=1}^{10} 1 = 10

Substituting these values into the sum expression:

S_{10} = \dfrac{1}{6} (2 \times 385 + 3 \times 55 + 10)

S_{10} = \dfrac{1}{6} (770 + 165 + 10)

S_{10} = \dfrac{945}{6} = \dfrac{315}{2}

Answer:

\dfrac{315}{2}

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