Question

prove by method of induction that
$5 + {5}^{2} + {5}^{3} + .... + {5}^{n} = \frac{5}{4}( {5}^{n} - 1)$
​

Answer 1

$\large\underline{\sf{Solution-}}$

To prove by Principal of Mathematical Induction,

$\rm \: 5 + {5}^{2} + {5}^{3} + - - - + {5}^{n} = \dfrac{5( {5}^{n} - 1)}{4}$

Let assume that

$\rm \: P(n) : 5 + {5}^{2} + {5}^{3} + - - - + {5}^{n} = \dfrac{5( {5}^{n} - 1)}{4}$

Step :- 1 For n = 1

$\rm \: P(1) : 5 = \dfrac{5( {5}^{1} - 1)}{4}$

$\rm \: P(1) : 5 = \dfrac{5(5 - 1)}{4}$

$\rm \: P(1) : 5 = \dfrac{5 \times 4}{4}$

$\rm \: P(1) : 5 = 5$

$\rm\implies \:P(n) \: is \: true \: for \: n = 1$

Step :- 2 For n = k, Assume that P(n) is true, where k is a natural number.

$\rm \: P(k) : 5 + {5}^{2} + {5}^{3} + - - - + {5}^{k} = \dfrac{5( {5}^{k} - 1)}{4}$

Step :- 3 For n = k + 1, We have to prove that P(n) is true.

That means, now we have to prove that

$\rm \: P(k + 1) : 5 + {5}^{2} + {5}^{3} + - - - + {5}^{k} + {5}^{k + 1} = \dfrac{5( {5}^{k + 1} - 1)}{4}$

Consider LHS

$\rm \: 5 + {5}^{2} + {5}^{3} + - - - + {5}^{k} + {5}^{k + 1}$

$\rm \: = \: (5 + {5}^{2} + {5}^{3} + - - - + {5}^{k}) + {5}^{k + 1}$

On substituting the value from step 2, we get

$\rm \: = \: \dfrac{5( {5}^{k} - 1) }{4} + {5}^{k + 1}$

$\rm \: = \: \dfrac{ {5}^{k + 1} - 5}{4} + {5}^{k + 1}$

$\rm \: = \: \dfrac{ {5}^{k + 1} - 5 + 4. {5}^{k + 1} }{4}$

$\rm \: = \: \dfrac{ {5}^{k + 1}(1 + 4) - 5}{4}$

$\rm \: = \: \dfrac{ {5}^{k + 1}(5) - 5}{4}$

$\rm \: = \: \dfrac{5( {5}^{k + 1} - 1)}{4}$

$\rm\implies \:P(n) \: is \: true \: for \: n = k + 1$

Hence,

By the process of Principal of Mathematical Induction,

$\boxed{\tt{ \: \: \rm \: 5 + {5}^{2} + {5}^{3} + - - - + {5}^{n} = \dfrac{5( {5}^{n} - 1)}{4} \: \: }}$