The general formula for the sum of successive odd squares from 1 to (2n-1) . It is supposed that the series end with square of an even number.
Sn = 12 + 22 + 32 +42
+ 5 2 + 62 + 72 + 82 +……… (n-1)2 +(n)2 = n (n+1) (2n+1)/6
The last number L = n ; Sn = L(L+1)(2L+1)/6
The given finite series is split into two series one with
odd squares and another with even squares.
12 + 22 + 32 +42
+ 5 2 + 62 + 72 + 82 +……… (n-1)2 +(n)2 = [12 + 32 + 5 2 + 72
+…… (2m-1)2 ]+ [ 22 +
42 + 62 + 82
+……… +(2m)2]
[12 + 32 + 5 2 + 72
+…… (2m-1)2 ] = [12
+ 22 + 32
+42 + 5 2 +
62 + 72 + 82
+……… (n-1)2 +(n)2]
– [22 + 42 + 62 + 82 +……… +(2m)2]
=
[n(n+1)(2n+1)/6] - 4 (n/2)[(n+2)/2] (n+1)/6 = n
(n2 – 1)/6
In terms of last number L
12 +
32 + 5 2 + 72 +…… (2m-1)2 ] = L(L+1)(2L+1)/6 –
L(L+1)(L+2)/6 = L(L2- 1)/6
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