Question

$1.2+{2.2}^2+{3.2}^3+\cdots +n.2^n=(n-1)2^{n+1}+2.$

Answer 1

Let P(n) = $1.2+{2.2}^2+{3.2}^3+\cdots +n.2^n=(n-1)2^{n+1}+2.$
For n = 1
$P\left(1\right)={1.2}^1=(1-1)2^{1+1}+2\Rightarrow 2=0+2\Rightarrow 2=2$
$\therefore$ P (1) is true
Let P(n) be true for n = k
$\therefore P\left(k\right)=1.2+{2.2}^2+{3.2}^2+\cdots +k{.2}^k=(k-1)2^{k+1}+2\text{For n}=k+1P(k+1)=1.2+{2.2}^2+{3.2}^3+\cdots +k.2^k+(k+1).2^{k+1}=(k+1-1)2^{k+1+1}+2=k.2^{k+1}+2+(k+1)2^{k+1}=2^{k+1}(k-1+k+1)+2=2^{k+1}\times 2k+2=k.2^{k+2}+2$
$\therefore P(k+1)$ is true
Thus P (k) is true $\Rightarrow P(k+1)$ is true
Hence by principle fo mathematical induction,
P(n) is true for al $nϵN.$