Question

The sum \( 1 + \frac{1}{2}(1^2+2^2) + \frac{1}{3}(1^2+2^2+3^2) + \ldots \) upto \(10\) terms is equal to:

• A. \(130\)
• B. \(155\)
• C. \( \frac{315}{2} \)
• D. \( \frac{325}{2} \)

Answer 1

The Correct Option is C

Concept: The sum of squares of first \(n\) natural numbers is: \[ 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6} \] Step 1: {Write the general term.} The \(k^{th}\) term is: \[ \frac{1}{k}(1^2+2^2+\cdots+k^2) \] Using the formula: \[ = \frac{1}{k}\cdot \frac{k(k+1)(2k+1)}{6} \] \[ = \frac{(k+1)(2k+1)}{6} \] Step 2: {Simplify the expression.} \[ (k+1)(2k+1)=2k^2+3k+1 \] Thus the term becomes: \[ \frac{2k^2+3k+1}{6} \] Step 3: {Sum from \(k=1\) to \(10\).} \[ S=\frac{1}{6}\sum_{k=1}^{10}(2k^2+3k+1) \] \[ =\frac{1}{6}\left(2\sum k^2+3\sum k+\sum 1\right) \] \[ \sum_{k=1}^{10}k=55, \qquad \sum_{k=1}^{10}k^2=385 \] \[ S=\frac{1}{6}(2\times385+3\times55+10) \] \[ =\frac{1}{6}(770+165+10) \] \[ =\frac{945}{6} \] \[ =\frac{315}{2} \]