Question

Find the sum of n terms of the series $3+8+22+72+266+1036+....$

• A. $\frac{3n}{4}+n(n+1)+\frac{1}{12}\left(4n-1\right)$
• B. $\frac{3n}{4}+n(n+1)+\frac{1}{12}\left(4^{n-1}\right)$
• C. $\frac{3n}{4}+n(n+1)+\frac{1}{12}\left(4n-2\right)$
• D. $\frac{3n}{4}+n(n+1)+\frac{1}{12}\left(4^{n-2}\right)$

Answer 1

The correct option is B $\frac{3n}{4}+n(n+1)+\frac{1}{12}\left(4^{n-1}\right)$

$S=3+8+22+72+266+1036+...$ ...(1)
$S=3+8+22+72+266+...$ ...(2)
From (1) - (2)
$0=3+5+14+50+194+770+...Tn$
$\Rightarrow Tn=3+5+14+50+194+770+...$ ...(3)
$Tn=3+5+14+50+194+770+...$ ...(4)
From (3) - (4)
$0=3+2+9+36+144+576+...$
$t_n=3+2+G.P$ with first term 9, common ratio 4, and number of terms n-2.
$\Rightarrow t_n=5+9\frac{({\left(4\right)}^{n-2}-1)}{4-1}\Rightarrow t_n=2+\frac{3}{16}\left(4^n\right)$
$\Rightarrow Tn=3+{\sum }_{n=1}^{n-1}\frac{3}{10}{4}^n$
$\Rightarrow Tn=3+2(n-1)+\frac{3}{16}\frac{\left(4\right)(4^{n-1}-1)}{4-1}$
$\Rightarrow Tn=\frac{3}{4}+2n+4^{(n-2)}$
$Sn=\sum Tn=\frac{3n}{4}+n(n+1)+\frac{1}{12}(4^n-1)$