Solve- 1^2-2^2+3^2-4^2+5^2-6^2......upto n terms
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HELLO DEAR,
If n is even, the sum is -n*(n+1)/2.
That is because, 1^2 - 2^2 = -1*(1+2) [ Since a^2 - b^2 = (a-b)*(a+b)
Also, 3^2 - 4^2 = -1*(3+4)
Thus , in the sum , taking -1 common, we have,
-1* (1+2+3+4+ .. + n) = -n*(n+1)/2
If the value of n is odd, the sum is clearly ,
S(n-1) + n^2
= -n*(n-1)/2 + n^2
= -n^2/2 + n^2 + n/2
= n^2/2 + n/2
=n*(n+1)/2
I HOPE ITS HELP YOU :-)
If n is even, the sum is -n*(n+1)/2.
That is because, 1^2 - 2^2 = -1*(1+2) [ Since a^2 - b^2 = (a-b)*(a+b)
Also, 3^2 - 4^2 = -1*(3+4)
Thus , in the sum , taking -1 common, we have,
-1* (1+2+3+4+ .. + n) = -n*(n+1)/2
If the value of n is odd, the sum is clearly ,
S(n-1) + n^2
= -n*(n-1)/2 + n^2
= -n^2/2 + n^2 + n/2
= n^2/2 + n/2
=n*(n+1)/2
I HOPE ITS HELP YOU :-)
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