Question

the sum of first n terms of the series 1square + 2* 2square + 3square +2*4square + 5square + 2*6square +... is n*(n+1)square/2 when n is even. What is the sum when is odd?

Answer 1

We know that the sum of
${1}^{2} + {2}^{2} + {3}^{2} + ..... + {n}^{2} = \frac{n(n + 1)(2n + 1)}{6}$
And
$2 \times {2}^{2} + 2 \times {4}^{2} + ....2 \times {(n)}^{2} \\ = 2( {1}^{2} + {2}^{2} + .... + {n}^{2} ) \\ = \frac{n(n + 1)(2n + 1)}{3}$
So u can find the value of
${1}^{2} + {3}^{2} + ..... {(2n - 1)}^{2}$
Finally, you will be able to solve your question!!
Thank me later!!☺☺

Answer 2

Answer:

Step-by-stWe know that the sum of

{1}^{2}  +  {2}^{2}  +  {3}^{2}  + .....  +  {n}^{2}  =  \frac{n(n + 1)(2n + 1)}{6}  

And

2 \times  {2}^{2}  + 2 \times  {4}^{2}  + ....2 \times  {(n)}^{2}  \\  = 2( {1}^{2}  +  {2}^{2}  + .... +  {n}^{2} ) \\  =  \frac{n(n + 1)(2n + 1)}{3}  

So u can find the value of

{1}^{2}  +  {3}^{2}  + ..... {(2n - 1)}^{2}  

Finally, you will be able to solve your question!!

Thank me later!!

ep explanation: