1.

Find the sum of n terms of series ` 2.4 + 6.8 + 10.12+.....`

Answer» when a=2,d=4
`a_n`=2+(n-1)4=4n-2
When a=4,d=4
`a_n`=4+(n-1)4=4n
`T_n=(4n-2)(4n)=8[2n^2-n)`
`S_n=sumT_n=sum16n^2-8n`
`(16n(n+1)(2n+1))/6-(8n(n+1))/2`
`(8*n(n+1))/2[(2(2n+1))/3-1]`
`4n(n+1)[(4n+2-3)/3]=(4n(n+1)(4n-1))/3`.


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