Question

If $1\cdot 3+3\cdot 3^2+5\cdot 3^3+7\cdot 3^4+\cdots$ up to $n$ terms is equal to $3+(n-1)3^t$, then $t$=

• A. $n-1$
• B. $2n-1$
• C. $n$
• D. $n+1$

Answer 1

Answer: (D)

$n+1$

Let $S=1.3+{3.3}^2+{5.3}^3+{7.3}^4+...+(2n-1)3^n\to \left(1\right)$
Multiplying (1) by $3$, we get
$3S=1.3^2+{3.3}^3+{5.3}^4+{7.3}^5+...+(2n-3)3^n+(2n-1)3^{n+1}\to \left(2\right)$
Subtracting (1) and (2) by shifting the terms, we get
$-2S=1.3+2(3^2+3^3+3^4+....+3^n)-(2n-1)3^{n+1}$
$\Rightarrow -2S=3+2\left[\frac{3^2(3^{n-1}-1)}{3-1}\right]-(2n-1)3^{n+1}$ [ sum of G.P series]
$\Rightarrow -2S=-6+23^{n+1}-2{n3}^{n+1}$
$\Rightarrow S=3+(n-1)3^{n+1}$
Therefore, $t=n+1$
Hence, option D.