Question

The sum of the series \( 1 + \frac{1}{2}(1^2 + 2^2) + \frac{1}{3}(1^2 + 2^2 + 3^2) + \dots \) up to 10 terms is:

• A. \(\frac{313}{2}\)
• B. \(\frac{315}{2}\)
• C. \(\frac{325}{2}\)
• D. \(\frac{335}{2}\)

Hint:

When the general term \(T_r\) is a polynomial in \(r\), the sum is easily calculated using the standard summation formulas for \(r^2\), \(r\), and constants.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We determine the general term \(T_r\) of the series. Each term is the average of the sum of the first \(r\) squares.

Step 2: Key Formula or Approach:
\[ T_r = \frac{1}{r} \sum_{k=1}^r k^2 = \frac{1}{r} \left[ \frac{r(r+1)(2r+1)}{6} \right] = \frac{(r+1)(2r+1)}{6} \] \[ T_r = \frac{2r^2 + 3r + 1}{6} \]

Step 3: Detailed Explanation:
We need the sum \(S_{10} = \sum_{r=1}^{10} T_r\): \[ S_{10} = \frac{1}{6} \left[ 2 \sum_{r=1}^{10} r^2 + 3 \sum_{r=1}^{10} r + \sum_{r=1}^{10} 1 \right] \] Using standard sums for \(n=10\): - \(\sum r^2 = \frac{10 \times 11 \times 21}{6} = 385\) - \(\sum r = \frac{10 \times 11}{2} = 55\) - \(\sum 1 = 10\) \[ S_{10} = \frac{1}{6} [ 2(385) + 3(55) + 10 ] = \frac{1}{6} [ 770 + 165 + 10 ] \] \[ S_{10} = \frac{945}{6} = \frac{315}{2} \]

Step 4: Final Answer:
The sum of the series up to 10 terms is \(\frac{315}{2}\).