2square +4square+6square+...........to n terms. find the sum of this series ?
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Answers
Answered by
25
Hey !!!
2² + 4² + 6².......2n² is even term
hence, 4 + 8 + 36......2n²
if we taking common as 4
then , 4 ( 1² + 2² + 3²......n² )
But , as we know that ,..sum of even square no.
1² + 2² + 3²......n² = n(n + 1 ) * (2n + 6)/6
hence,
4n( n + 1 ) × ( 2n + 6 ) /6
=> 2n ( n + 1 ) ( 2n + 6) / 3 Answer ✔
___________________________
Hope it helps you !!! ..
@Rajukumar111
2² + 4² + 6².......2n² is even term
hence, 4 + 8 + 36......2n²
if we taking common as 4
then , 4 ( 1² + 2² + 3²......n² )
But , as we know that ,..sum of even square no.
1² + 2² + 3²......n² = n(n + 1 ) * (2n + 6)/6
hence,
4n( n + 1 ) × ( 2n + 6 ) /6
=> 2n ( n + 1 ) ( 2n + 6) / 3 Answer ✔
___________________________
Hope it helps you !!! ..
@Rajukumar111
Answered by
26
☜☆☞hey friend ☜☆☞
here is your answer ☞
_________
2² + 4² + 6².......2n² is series
then, 4 + 8 + 36......2n²
we take a common no. as 4
we get, 4 ( 1² + 2² + 3²......n² )
we know that ,..
1² + 2² + 3²......n² = [n(n + 1 )×(2n + 6)]/6
hence,
[4n( n + 1 ) × ( 2n + 6 )] /6
[ 2n ( n + 1 ) ( 2n + 6) ]/ 3 ___answer
hope it will help you
here is your answer ☞
_________
2² + 4² + 6².......2n² is series
then, 4 + 8 + 36......2n²
we take a common no. as 4
we get, 4 ( 1² + 2² + 3²......n² )
we know that ,..
1² + 2² + 3²......n² = [n(n + 1 )×(2n + 6)]/6
hence,
[4n( n + 1 ) × ( 2n + 6 )] /6
[ 2n ( n + 1 ) ( 2n + 6) ]/ 3 ___answer
hope it will help you
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