Question

$1.3+{\left(2.3\right)}^2+(3.3{)}^3+....+(n.3{)}^n=\frac{\left(2n-1\right)3^{n+1}+3}{4}$

• A. True
• B. False

Answer 1

The correct option is A True

To prove or disprove

$1.3+(2.3{)}^2+(3.3{)}^3+.....+(n.3{)}^n=\frac{(2n-1)3^{n+1}+3}{4}$

$P\left(n\right)=1.3+(2.3{)}^2+(3.3{)}^3+......+(n.3{)}^n=\frac{(2n-1)3^{n+1}+3}{4}$

Let $n=1$,

$P\left(1\right)=1.3=\frac{(2\times 1-1)3^{1+1}+3}{4}$

$\because 1.3=3=LHS$
$\frac{1\times 3^2+3}{4}=\frac{9+3}{4}=\frac{12}{4}=3=RHS$

$\therefore LHS=RHS$

$P\left(n\right)$ is true for $n=1$

Assume $P\left(k\right)$ is true,

i.e, $\left(1.3\right)+(2.3{)}^2+(3.3{)}^3+.......+(k.3{)}^k=\frac{(2k-1)3^{k+1}+3}{4}$ ...(1)

Now let us prove that $P(k+1)$ is true,

LHS: $1.3+(2.3{)}^2+(3.3{)}^3+...(k.3{)}^k+[(k+1).3{]}^{k+1}$

substitute eq (1),

$=\frac{(2k-1)3^{k+1}+3}{4}+\left[\right(k+1).3{]}^{k+1}$

$=\frac{(2k-1)3^{k+1}+3+(4k+4)\cdot 3^{k+1}}{4}$

$=\frac{(6k+3)3^{k+1}+3}{4}=\frac{3\times (2k+1)3^{k+1}+3}{4}$

$=\frac{(2k+1)3^{k+1+1}+3}{4}=\frac{\left[2\right(k+1)-1]3^{(k+1)+1}+3}{4}$

RHS: $\frac{\left[2\right(k+1)-1]3^{(k+1)+1}+3}{4}$

LHS = RHS when $P\left(k\right)$ is true.

hence proved.