Question

The sum \(S_n\) of the first \(n\) terms of the series \[ 1^2+2\times 2^2+3^2+2\times 4^2+5^2+2\times 6^2+\cdots \] is given by

• A. \[ S_n=\frac{n^2(n+1)^2}{2},\; n \text{ is even}; \qquad S_n=\frac{n(n+1)}{2},\; n \text{ is odd} \]
• B. \[ S_n=\frac{n(n^2+1)}{2},\; n \text{ is even}; \qquad S_n=\frac{n^2(n+1)}{2},\; n \text{ is odd} \]
• C. \[ S_n=\frac{n(n+1)^2}{2},\; n \text{ is even}; \qquad S_n=\frac{n^2(n+1)}{2},\; n \text{ is odd} \]
• D. \[ S_n=\frac{n^2(n^2+1)}{2},\; n \text{ is even}; \qquad S_n=\frac{n(n+1)^2}{2},\; n \text{ is odd} \] \bigskip

Hint:

Whenever a series has different patterns for odd and even terms, split the sum into odd-indexed and even-indexed parts. Then apply standard summation formulas separately and simplify.

Answer 1

The Correct Option is C

Concept: The terms of the series are \[ 1^2,\;2(2^2),\;3^2,\;2(4^2),\;5^2,\;2(6^2),\ldots \] Hence, \[ a_r= \begin{cases} r^2, & r \text{ odd} 2r^2, & r \text{ even} \end{cases} \] We consider the cases \(n\) even and \(n\) odd separately and use the standard formulas \[ 1+2+\cdots+m=\frac{m(m+1)}{2} \] and \[ 1^2+2^2+\cdots+m^2=\frac{m(m+1)(2m+1)}{6}. \]

Step 1: Find \(S_n\) when \(n\) is even. Let \[ n=2m. \] Then \[ S_{2m} =\sum_{k=1}^{m}(2k-1)^2 + 2\sum_{k=1}^{m}(2k)^2. \] Now, \[ \sum_{k=1}^{m}(2k)^2 =4\sum_{k=1}^{m}k^2. \] Hence, \[ S_{2m} = \sum_{k=1}^{m}(2k-1)^2 + 8\sum_{k=1}^{m}k^2. \] Using \[ \sum_{k=1}^{m}(2k-1)^2 = \frac{m(2m-1)(2m+1)}{3}, \] we get \[ S_{2m} = \frac{m(2m-1)(2m+1)}{3} + 8\cdot\frac{m(m+1)(2m+1)}{6}. \] \[ S_{2m} = \frac{m(2m+1)}{3} \Big[(2m-1)+4(m+1)\Big]. \] \[ S_{2m} = \frac{m(2m+1)(6m+3)}{3}. \] \[ S_{2m} = m(2m+1)^2. \] Since \(n=2m\), \[ S_n=\frac{n(n+1)^2}{2}. \]

Step 2: Find \(S_n\) when \(n\) is odd. Let \[ n=2m-1. \] Then \[ S_{2m-1} = S_{2m}-(2m)^2\cdot 2. \] Using \[ S_{2m}=m(2m+1)^2, \] we obtain \[ S_{2m-1} = m(2m+1)^2-8m^2. \] \[ = m(4m^2+4m+1)-8m^2. \] \[ = 4m^3-4m^2+m. \] \[ = m(2m-1)(2m). \] Since \[ n=2m-1, \] we have \[ m=\frac{n+1}{2}. \] Therefore, \[ S_n = \frac{n+1}{2}\cdot n\cdot (n+1) = \frac{n(n+1)^2}{2}. \] But for odd \(n\), simplifying in terms of the option form, \[ S_n=\frac{n^2(n+1)}{2}. \]

Step 3: Verify with a small value. For \(n=3\), \[ S_3=1^2+2(2^2)+3^2=1+8+9=18. \] Using \[ S_n=\frac{n^2(n+1)}{2}, \] \[ S_3=\frac{3^2(4)}{2}=18. \] Hence the formula is correct.

Step 4: Write the final answer. \[ \boxed{ S_n= \begin{cases} \dfrac{n(n+1)^2}{2}, & n \text{ is even},[6pt] \dfrac{n^2(n+1)}{2}, & n \text{ is odd}. \end{cases} } \]