Find the sum of the first 2n terms of the series 1^2+2+3^2+4+5^2+6....
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12
2^2 - 1^2 = (2 + 1) (2 - 1) = 3 * 1 = 3
4^2 - 3^2 = (4 + 3) (4 - 3) = 7 * 1 = 7
6^2 - 5^2 = (6 + 5) (6 - 5) = 11 * 1 = 11
It's an arithmetic series with difference = 4
2n terms = n pairs
n
1 ... 2 ... 3 .... 4 ... .... n
3 ... 7 ...11 ... 15 .... (4n-1)
Using formula for sum of arithmetic series
(first + last) * (1/2 the number of terms)
Sum = (3 + (4n-1)) * n / 2
or
(4n + 2) * n / 2
or
(2n + 1) * n
4^2 - 3^2 = (4 + 3) (4 - 3) = 7 * 1 = 7
6^2 - 5^2 = (6 + 5) (6 - 5) = 11 * 1 = 11
It's an arithmetic series with difference = 4
2n terms = n pairs
n
1 ... 2 ... 3 .... 4 ... .... n
3 ... 7 ...11 ... 15 .... (4n-1)
Using formula for sum of arithmetic series
(first + last) * (1/2 the number of terms)
Sum = (3 + (4n-1)) * n / 2
or
(4n + 2) * n / 2
or
(2n + 1) * n
Answered by
10
there is your answer
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