Sum of n terms of series `12+16+24+40+....` (A) `2(2^n -1)+8n` (B) `2(2^n-1)+6n` (C) `3(2^n-1)+8n` (D) `4(2^n-1)+8n`

Answer

N/a

Explanation

In the given series, the differences of consecutive terms (16 12), (24-16), (40-24), …
`=4, 8, 16, …` form a G.P. Therefore, `T_(n)` can be determined by difference method.
Let nth terms and sum of n terms of the series be `T_(n)" and "S_(n)` respectively.
`therefore" "S_(n)=12+16+24+40...+T_(n-1)+T_(n)`
`underline(S_(n)=" "12+16+24+...+T_(n-1)+T_(n))`
On subtracting
`0=[12+4+8+16+...+t_(n)]-T_(n)`
`rArr" "T_(n)=12+{4+8+16+..."upto (n-1) terms"}`
`=12+(4cdot(2^(n-1)-1))/(2-1)`
`=12+2^(n+1)-4=2^(n+1)+8`
`rArr" "S_(n)=Sigma2^(n+1)+8n`
`=(2^(2)+2^(3)+2^(4)+...+2^(n+1))=8n`
`(2^(2)(2^(n)-1))/(2-1)+8n`
`=2^(n+2)-4+8n=(2^(n+2)+8n-4).`