Find the sum of series: 1^2 - 2^2 + 3^2 - 4^2 + 5^2....

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cooldudeachyut

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Question : Find the sum of the series - 12 - 22 + 32 - 42 + 52 ...... n terms

My attempt : I took two terms of the series as (2k-1)2 - (2k)2 = 1-4k

So adding the series as ∑(1-4k) taking range of k from 1 to n/2 (since I'm taking two terms at the same time), I got the answer :- -n(n+1)/2
but my textbook has given the answer as :- -n(2n+1)

Please help.
 
cooldudeachyut said:
Question : Find the sum of the series - 12 - 22 + 32 - 42 + 52 ...... n terms

My attempt : I took two terms of the series as (2k-1)2 - (2k)2 = 1-4k

So adding the series as ∑(1-4k) taking range of k from 1 to n/2 (since I'm taking two terms at the same time), I got the answer :- -n(n+1)/2
but my textbook has given the answer as :- -n(2n+1)

Please help.
Click to expand...
n may be either even, n=2j, or odd, n=2j-1. For n=2j, the sum will end on the minus sign, i.e.
S(n) = S(2j) = 12 - 22 + 32 - 42 ... + (2j-1)2 - (2j)2
=(1-4*1) + (1-4*2) + ... + (1-4j)
= j - 4 j (j+1) / 2
= j - 2 j (j+1) = j (1-2j-2) = -j (2j+1) = -2j (2j+1)/2
= -n (n+1)/2
So it looks like you are correct for n even. However, for n odd ...
 
Ishuda said:
n may be either even, n=2j, or odd, n=2j-1. For n=2j, the sum will end on the minus sign, i.e.
S(n) = S(2j) = 12 - 22 + 32 - 42 ... + (2j-1)2 - (2j)2
=(1-4*1) + (1-4*2) + ... + (1-4j)
= j - 4 j (j+1) / 2
= j - 2 j (j+1) = j (1-2j-2) = -j (2j+1) = -2j (2j+1)/2
= -n (n+1)/2
So it looks like you are correct for n even. However, for n odd ...
Click to expand...

for n = 2j+1

Sum = (sum for 2j) + (2j+1)^2

= -2j (2j+1)/2 + (2j+1)^2

=(2j+1)[(2j+1) - j]/2

=(2j+1)[(j+1) ]

=(2j+1)[(2j+1) + 1 ]/2

= n(n+1)/2................This has a positive sign
 
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