Introduction to P.M.I
Question

# The sum of series $1\cdot 3+2\cdot {3}^{2}+3\cdot {3}^{3}+\dots +\mathrm{n}\cdot {3}^{\mathrm{n}}$ is

Moderate
Solution

## Let the given statement be P(n). $\mathrm{P}\left(\mathrm{n}\right):1\cdot 3+2\cdot {3}^{2}+3\cdot {3}^{3}+\dots +\mathrm{n}\cdot {3}^{\mathrm{n}}=\frac{\left(2\mathrm{n}-1\right){3}^{\mathrm{n}+1}+3}{4}$Step l: Let it is true for n = 1,      $\mathrm{P}\left(1\right):\frac{\left(2\cdot 1-1\right){3}^{1+1}+3}{4}=\frac{{3}^{2}+3}{4}=\frac{9+3}{4}=\frac{12}{4}$    $⇒3=1.3$  which is trueStep II: For n=k, Step III: For n=k+1,On taking 3k+1 common in first and last term of numerator part, $=\frac{{3}^{\mathrm{k}+1}\left(2\mathrm{k}-1+4\mathrm{k}+4\right)+3}{4}=\frac{{3}^{\mathrm{k}+1}\left(6\mathrm{k}+3\right)+3}{4}$On taking 3 common in first term of numerator part,$\begin{array}{l}=\frac{{3}^{\mathrm{k}+1}\cdot 3\left(2\mathrm{k}+1\right)+3}{4}\\ =\frac{{3}^{\left(\mathrm{k}+1\right)+1}\left(2\mathrm{k}+2-1\right)+3}{4}\\ =\frac{\left\{2\left(\mathrm{k}+1\right)-1\right\}{3}^{\left(\mathrm{k}+1\right)+1}+3}{4}\end{array}$Therefore, P(k + 1) is true when P(k) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.

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