Question

If $1.3+{2.3}^2+{3.3}^3+...+n{.3}^n$= $\frac{(2n-1)3^a+b}{4}$ then $a$ and $b$ are respectively

• A. $n,2$
• B. $n,3$
• C. $n+1,2$
• D. $n+1,3$

Answer 1

Answer: (D)

$n+1,3$

$1.3+{2.3}^2+{3.3}^3+...+n{.3}^n$

this can be wriiten as ${0.3}^0+1.3+{2.3}^2+{3.3}^3+...+n{.3}^n$

Sum of $n$ terms in AGP is $S_n=\frac{a-[a+(n-1\left)d\right]r^n}{1-r}+\frac{dr(1-r^{n-1})}{(1-r{)}^2}$

Here $a=0,d=1,n=n,r=3$

Substituting in the given formula we get,

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

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

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

Thus, $a$ and $b$ are $n+1$ and $3$ respectively.