Question

Prove that $\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{\cdot }\mathbf{5}\mathbf{+}\mathbf{3}\mathbf{\cdot }{\mathbf{5}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}\mathbf{3}\mathbf{\cdot }{\mathbf{5}}^{\mathbf{n}}\mathbf{=}\mathbf{3}\left(5^{n+1}-1\right)\mathbf{/}\mathbf{4}$whenever nis a nonnegative integer.

Short answer

It is proved that $3+3\cdot 5+3\cdot 5^2+\dots +3\cdot 5^n=3\left(5^{n+1}-1\right)/4$ whenever nis a nonnegative integer.

Step-by-step solution

Principle of Mathematical Induction

To prove that P(n)is true for all positive integers n, where p(n)is a propositional function, we complete two steps:

Basis Step:

We verify that P(1)is true.

Inductive Step:

We show that the conditional statement $\mathit{p}\left(k\right)\mathbf{\to }\mathit{p}\left(k+1\right)$is true for all positive integers k.

Proving the basis step

Given statement is

$3+3\cdot 5+3\cdot 5^2+\dots +3\cdot 5^n=\frac{3\left(5^{n+1}-1\right)}{4}$

In the basis step we need to prove that P(1) is true

For finding statement p(1) substituting 1 for n in the statement

Therefore, the statement P(1) is

$\begin{array}{r}3+3\cdot 5^1=\frac{3\left(5^{1+1}-1\right)}{4}\\ 18=\frac{3(25-1)}{4}\\ 18=\frac{3\cdot 24}{4}\\ 18=18\end{array}$

Therefore, the statement P(1) is true this is also known as the basis step of the proof.

Proving the Inductive step

In the inductive step, we need to prove that, if P(k) is true, then P(k+1) is also true.

That is,

$P\left(k\right)\to P\left(k+1\right)$is true for all positive integers k.

In the inductive hypothesis, we assume that P(k) is true for any arbitrary positive integer

$3+3\cdot 5+3\cdot 5^2+\dots +3\cdot 5^k=\frac{3\left(5^{k+1}-1\right)}{4}$...(i)

Now we must have to show that is also true

Therefore replacing with in the statement

$\begin{array}{r}3+3\cdot 5+3\cdot 5^2+\dots +3\cdot 5^{(k+1)}=\frac{3\left(5^{k+1+1}-1\right)}{4}\\ =\frac{3\left(5^{k+2}-1\right)}{4}\end{array}$

Now, Adding $3.5^{\left(K+1\right)}$in both sides of the equation (i) or inductive hypothesis.