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the sum of first n terms of the series 12+2.22+32+2.42+52+2.62... is n(n+1)2/2, where n is even. when n is odd, the sum is...?

Samuel Garry , 7 Years ago
Grade 11
anser 1 Answers
Arun
When n is odd, let n = 2m +1
∴ The req. sum
= 12 + 2.22 + 32 + 2.42 + ….. + 2(2m) 2 + (2m + 1) 2
= Σ (2m + 1) 2 + 4[12 + 22 + 32 + …. +m2 ]
=(2m + 1) (2m + 2) (4m + 2 + 1)/6 + 4m (m + 1) (2m + 1)/6
= (2m + 1) (m + 1)/6 [2 (4m + 3) + 4m ]
= (2m + 1) (2m + 2)(6m + 3)/6 = ((2m + 1)2(2m + 2)/2
= n2(n + 1)/2 [ ∵ 2m + 1 =n]
ALTERNATE SOLUTION:
∵ n is odd, last term = n2
∴ req. sum
= [12 + 2.22 + 32 + 2.42 +….. + 2(n-1) 2 ] + n2
= (n – 1)n2 /2 +n2 [ Using sum for( n – 1 ) to be even = n2 (n + 1)/2
Last Activity: 7 Years ago
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