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

Step 1: Understanding the Question
We are asked to find the sum of the first 10 terms of a given series. The general term of the series involves the sum of the squares of the first \(k\) natural numbers.
Step 2: Key Formula or Approach
The sum of the squares of the first \(n\) natural numbers is given by: \[ \sum_{i=1}^{n} i^2 = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6} \] We will also need the formulas for the sum of the first \(n\) natural numbers and the sum of a constant: \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \quad \text{and} \quad \sum_{k=1}^{n} 1 = n \] Step 3: Detailed Explanation
First, let's find the expression for the \(k^{th}\) term of the series, denoted by \(T_k\).
\[ T_k = \frac{1}{k} (1^2 + 2^2 + 3^2 + \dots + k^2) \] Using the formula for the sum of squares: \[ T_k = \frac{1}{k} \left[ \frac{k(k+1)(2k+1)}{6} \right] \] Canceling \(k\) from the numerator and denominator: \[ T_k = \frac{(k+1)(2k+1)}{6} \] Expanding the numerator: \[ T_k = \frac{2k^2 + 3k + 1}{6} \] Now, we need to find the sum of the first 10 terms, which is \( S_{10} = \sum_{k=1}^{10} T_k \).
\[ S_{10} = \sum_{k=1}^{10} \frac{2k^2 + 3k + 1}{6} \] \[ S_{10} = \frac{1}{6} \left[ \sum_{k=1}^{10} (2k^2 + 3k + 1) \right] \] \[ S_{10} = \frac{1}{6} \left[ 2 \sum_{k=1}^{10} k^2 + 3 \sum_{k=1}^{10} k + \sum_{k=1}^{10} 1 \right] \] Now, we use the standard summation formulas for \(n=10\):
\[ \sum_{k=1}^{10} k^2 = \frac{10(10+1)(2 \cdot 10+1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = 5 \cdot 11 \cdot 7 = 385 \] \[ \sum_{k=1}^{10} k = \frac{10(10+1)}{2} = \frac{10 \cdot 11}{2} = 55 \] \[ \sum_{k=1}^{10} 1 = 10 \] Substitute these values back into the expression for \(S_{10}\):
\[ S_{10} = \frac{1}{6} [ 2(385) + 3(55) + 10 ] \] \[ S_{10} = \frac{1}{6} [ 770 + 165 + 10 ] \] \[ S_{10} = \frac{1}{6} [ 945 ] \] \[ S_{10} = \frac{945}{6} = \frac{315}{2} \] Step 4: Final Answer
The sum of the first 10 terms is \( \frac{315}{2} \).