Prove the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + n \cdot 3^{n} = \frac{(2n - 1) 3^{n+1} + 3}{4}$
$1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + n \cdot 3^{n} = \frac{(2n - 1) 3^{n+1} + 3}{4}$
Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + n \cdot 3^{n} = \frac{(2n - 1) 3^{n+1} + 3}{4}$
For $n = 1$,we have
$P(1): 1 \cdot 3 = 3 = \frac{(2 \cdot 1 - 1) 3^{1+1} + 3}{4} = \frac{3^{2} + 3}{4} = \frac{12}{4} = 3$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + k \cdot 3^{k} = \frac{(2k - 1) 3^{k+1} + 3}{4}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$(1 \cdot 3 + 2 \cdot 3^{2} + \ldots + k \cdot 3^{k}) + (k+1) \cdot 3^{k+1}$
$= \frac{(2k - 1) 3^{k+1} + 3}{4} + (k+1) 3^{k+1}$ [Using $(i)$]
$= \frac{(2k - 1) 3^{k+1} + 3 + 4(k+1) 3^{k+1}}{4}$
$= \frac{3^{k+1} \{2k - 1 + 4k + 4\} + 3}{4}$
$= \frac{3^{k+1} \{6k + 3\} + 3}{4}$
$= \frac{3^{k+1} \cdot 3 \{2k + 1\} + 3}{4}$
$= \frac{3^{(k+1)+1} \{2(k+1) - 1\} + 3}{4}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.
$P(n): 1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + n \cdot 3^{n} = \frac{(2n - 1) 3^{n+1} + 3}{4}$
For $n = 1$,we have
$P(1): 1 \cdot 3 = 3 = \frac{(2 \cdot 1 - 1) 3^{1+1} + 3}{4} = \frac{3^{2} + 3}{4} = \frac{12}{4} = 3$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + k \cdot 3^{k} = \frac{(2k - 1) 3^{k+1} + 3}{4}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$(1 \cdot 3 + 2 \cdot 3^{2} + \ldots + k \cdot 3^{k}) + (k+1) \cdot 3^{k+1}$
$= \frac{(2k - 1) 3^{k+1} + 3}{4} + (k+1) 3^{k+1}$ [Using $(i)$]
$= \frac{(2k - 1) 3^{k+1} + 3 + 4(k+1) 3^{k+1}}{4}$
$= \frac{3^{k+1} \{2k - 1 + 4k + 4\} + 3}{4}$
$= \frac{3^{k+1} \{6k + 3\} + 3}{4}$
$= \frac{3^{k+1} \cdot 3 \{2k + 1\} + 3}{4}$
$= \frac{3^{(k+1)+1} \{2(k+1) - 1\} + 3}{4}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.
Explore More
Similar Questions
Prove the statement by the Principle of Mathematical Induction: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$.
Difficult
View SolutionProve the following by using the principle of mathematical induction for all $n \in N$:
$a + ar + ar^{2} + \ldots + ar^{n-1} = \frac{a(r^{n} - 1)}{r - 1}$
$a + ar + ar^{2} + \ldots + ar^{n-1} = \frac{a(r^{n} - 1)}{r - 1}$
Medium
View SolutionUse the Principle of Mathematical Induction to prove that $\cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \cos 2^{n-1} \theta = \frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$ for all $n \in N$.
Difficult
View SolutionIf $P(n) = 2 + 4 + 6 + \dots + 2n$,$n \in N$,then $P(k) = k(k + 1) + 2 \implies P(k + 1) = (k + 1)(k + 2) + 2$ for all $k \in N$. So we can conclude that $P(k) = k(k + 1) + 2$ for all $k \in N$ is true. What can we conclude about $P(n) = n(n + 1) + 2$ for all $n \in N$?
Easy
View SolutionFor every natural number $n$,which of the following inequalities is true?
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo