First slide

Introduction to P.M.I

Question

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

Moderate

A

$\frac{(2 n + 1) 3^{n + 1} + 3}{4}$
B

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

$\frac{(2 n - 1) 3^{n - 1} - 3}{4}$
D

None of these

Solution

Let the given statement be P(n).
$P (n) : 1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \dots + n \cdot 3^{n} = \frac{(2 n - 1) 3^{n + 1} + 3}{4}$
Step l: Let it is true for n = 1,
$P (1) : \frac{(2 \cdot 1 - 1) 3^{1 + 1} + 3}{4} = \frac{3^{2} + 3}{4} = \frac{9 + 3}{4} = \frac{12}{4}$
$\Rightarrow 3 = 1.3$ which is true
Step II: For n=k,
$i.e. 1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \dots + k \cdot 3^{k} = \frac{(2 k - 1) 3^{k + 1} + 3}{4} - - - - - (i)$
Step III: For n=k+1,
$\begin{array}{l} (1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \dots + k \cdot 3^{k}) + (k + 1) 3^{k + 1} \\ = \frac{(2 k - 1) 3^{k + 1} + 3}{4} + (k + 1) 3^{k + 1} [using Eq. (i)] \\ = \frac{(2 k - 1) 3^{k + 1} + 3 + 4 (k + 1) 3^{k + 1}}{4} \end{array}$
On taking 3^k+1 common in first and last term of numerator part,
$= \frac{3^{k + 1} (2 k - 1 + 4 k + 4) + 3}{4} = \frac{3^{k + 1} (6 k + 3) + 3}{4}$
On taking 3 common in first term of numerator part,
$\begin{array}{l} = \frac{3^{k + 1} \cdot 3 (2 k + 1) + 3}{4} \\ = \frac{3^{(k + 1) + 1} (2 k + 2 - 1) + 3}{4} \\ = \frac{{2 (k + 1) - 1} 3^{(k + 1) + 1} + 3}{4} \end{array}$
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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