Question

The sum of the $n$ terms of the series $3+8+22+72+266+1036+\cdots$

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

Answer 1

The correct option is A $n(n+1)+\frac{1}{3}(4^n-1)$

Given series is $3,8,22,72,266,1036,\cdots$
So first order difference is
$5,14,50,194,770,\cdots$
And second order difference is
$9,36,144,576,\cdots$
As we can observe that second order difference is in G.P.
$\therefore T_n=a{4}^{n-1}+bn+c{T}_1=3=a+b+c\cdots \left(i\right)T_2=8=4a+2b+c\cdots \left(ii\right)T_3=22=16a+3b+c\cdots \left(iii\right)$
from $\left(ii\right)-\left(i\right)$ and from $\left(iii\right)-\left(i\right)$
$3a+b=5\cdots \left(iv\right)15a+2b=19\cdots \left(v\right)$
from $\left(v\right)-2\left(iv\right)$
$a=1,b=2,c=0$
hence $T_n=4^{n-1}+2n$

$\therefore S_n=\sum T_n=2\sum n+\sum 4^{n-1}=n(n+1)+\frac{1}{3}(4^n-1)$

Alternate Solution:
Substitute $n=1,2,3\cdots$ in the options given
and verify it with the given series