Question

Sum of \( 1 + \dfrac{1}{2}(1^2 + 2^2) + \dfrac{1}{3}(1^2 + 2^2 + 3^2) + \dots \) up to 10 terms is

• A. \( \dfrac{313}{2} \)
• B. \( \dfrac{315}{2} \)
• C. 313
• D. 315

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We first identify the general term \( T_r \) of the series and then apply the summation formula for the first 10 terms.
Step 2: Key Formula or Approach:
The \( r \)-th term is given by: \[ T_r = \frac{1}{r} \sum_{k=1}^{r} k^2 \] Using the sum of squares formula \( \sum_{k=1}^{r} k^2 = \frac{r(r+1)(2r+1)}{6} \): \[ T_r = \frac{1}{r} \cdot \frac{r(r+1)(2r+1)}{6} = \frac{(r+1)(2r+1)}{6} = \frac{2r^2 + 3r + 1}{6} \]
Step 3: Detailed Explanation:
The sum of 10 terms is \( S_{10} = \sum_{r=1}^{10} T_r \): \[ S_{10} = \sum_{r=1}^{10} \frac{2r^2 + 3r + 1}{6} = \frac{1}{6} \left[ 2 \sum r^2 + 3 \sum r + \sum 1 \right] \] Using standard summation results for \( n=10 \): - \( \sum_{r=1}^{10} r^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \) - \( \sum_{r=1}^{10} r = \frac{10 \cdot 11}{2} = 55 \) - \( \sum_{r=1}^{10} 1 = 10 \) \[ S_{10} = \frac{1}{6} [ 2(385) + 3(55) + 10 ] = \frac{1}{6} [ 770 + 165 + 10 ] \] \[ S_{10} = \frac{945}{6} = \frac{315}{2} \]
Step 4: Final Answer:
The sum up to 10 terms is \( \dfrac{315}{2} \).