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Prove the following by using the principle a mathematical induction for all $n\epsilon N$.
$1+3+3^{2}+\ldots+3^{n-1}=\frac{\left(3^{n}-1\right)}{2}$
Given: $1+3+3^{2}+\ldots+3^{n-1}=\frac{\left(3^{n}-1\right)}{2}$
To do: To prove that $1+3+3^{2}+\ldots+3^{n-1}=\frac{\left(3^{n}-1\right)}{2}$ by using the principle of mathematical induction for all $n\epsilon N$
Solution:
Let us assume $P ( n):1+3+3^2++3^{n-1}=\frac{( 3^n-1)}{2}$
If $n=1,\ L.H.S.=1$
$R.H.S.=\frac{( 3^1-1)}{2}=\frac{( 3^1)}{2}=\frac{( 2)}{2}=1$
Thus, $L.H.S.=R.H.S.$
$P( n)$ is true for $n=1$
We assume that $P( k)$ is true
$1+3+3^2+..+ 3^{k-1}=\frac{( 3^k-1)}{2}$ ............$( i)$
For $P( k+1)$:
$1+3+3^2+..+ 3^{( k+1)-1}=\frac{( 3^{k+1}-1)}{2}$
$1+3+3^2+...3^{k-1}+3^k=\frac{( 3^{k+1}-1)}{2}$ ........$( ii)$
We should prove $P( k+1)$ from $P( k)$ i.e. $( ii)$ from $( i)$
From $( i)$,
$1+3+3^2+..+ 3^{k-1}=\frac{( 3^k-1)}{2}$
Adding $3^k$ both sides
$1+3+3^2+..+3^{k-1}+3^k=\frac{( 3^k-1)}{2}+3^k$
$=\frac{( 3^k-1)+2( 3^k))}{2}$
$=\frac{( 3^k-1+2(3^k)}{2}$
$=\frac{( 3(3^k)-1)}{2}$
$=\frac{( 3^{k+1}-1)}{2}$
Thus, $1+3+3^2+..3^{k-1}+3^{k}=\frac{( 3^{k+1})-1)}{2}$
Thus we find that $P( k+1)$ is always true when $P( k)$ is true By the principle of mathematical induction,
$P( n)$ is true for $n\epsilon N$
Updated on: 10-Oct-2022
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