The sum of the first n terms of the series\\ 1^2 + 2 2^2 + 3^2 + 2 4^2 + 5^2 + 2 6^2 + is n(n + 1)^22, when n is even. when n is odd, the sum is

Given 1^2+2 2^2+3^2+2 4^2+5^2+2 6^2+. to n terms=n(n+1)^22 n is even. to find the sum when n is odd. have S_2 m=1^2+2 2^2+3^2+2 4^2+ +2(2 m)^2aligned& =(2 m)(2 m+1)^22=(2 m+1)^2(m+1) \\& =n^2(n+12)( n=m+1 m=n-12)aligned n be odd, n=2 m+1aligned_2 m+1 & =S_2 m+T_2 m+1 \\& =1^2+2 2^2+3^2+2 4^2++2(2 m)^2+(2 m+1)^2 \\& =(2 m)(2 m+1)^22+(2 m+1)^2=(2 m+1

The sum of the first n terms of the series (1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2 . 6^{2} + ..) is (\frac{n(n + 1)^{2}}{2}, when n is even. when n is odd, the sum is: Sequence & Series (Mathematics)

Given

1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+2 \cdot 6^2+\cdots

. to

n

terms\n

\n=\frac{n(n+1)^2}{2}\n

\n \nWhen

n

is even.\nRequired to find the sum when

n

is odd.\nWe have

S_{2 m}=1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+\cdots \cdot+2(2 m)^2

\n\begin{aligned}\n& =\frac{(2 m)(2 m+1)^2}{2}=(2 m+1)^2(m+1) \\\n& =n^2\left(\frac{n+1}{2}\right)\left(\because n=m+1 \Rightarrow m=\frac{n-1}{2}\right)\n\end{aligned}\n

\n \nLet

n

be odd,

n=2 m+1

\n\begin{aligned}\nS_{2 m+1} & =S_{2 m}+T_{2 m+1} \\\n& =1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+\cdots+2(2 m)^2+(2 m+1)^2 \\\n& =\frac{(2 m)(2 m+1)^2}{2}+(2 m+1)^2=(2 m+1)^2(m+1) \\\n=n^2 & \left(\frac{n+1}{2}\right)\left(\because n=m+1 \Rightarrow m=\frac{n-1}{2}\right)\n\end{aligned}\n

Explanation

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