Chapter 12.5, Problem 5E

To determine

To Prove: $5+8+11+\mathrm{.....}+\left(3n+2\right)=\frac{n\left(3n+7\right)}{2}$ is True for all the natural numbers $n$ .

Explanation of Solution

Given information:

Use Mathematical induction to prove the above mentioned formula.

Concept used:

1.By using Mathematical induction technique we need to prove that this formula holds True for all natural numbers. The first step is to prove True for $n=1$ .

2.Second step by using induction hypothesis of mathematical induction we assume for $n=k$ given formula is True.

3.Third step is we need to prove that above formula is True for $n=k+1$ by using formula is True for $n=k$ .

Proof:

First we need to prove the above formula is True for $n=1$ .

Given formula is

  $5+8+11+\mathrm{.....}+\left(3n+2\right)=\frac{n\left(3n+7\right)}{2}$

Put $n=1$

We get

Left hand side = 5

Right hand side = $\frac{n\left(3n+7\right)}{2}=\frac{1\left(3+7\right)}{2}=\frac{10}{2}=5$

So Left hand side = Right hand side

It True for $n=1$ .

Let us assume it is True for $n=k$

So

  $5+8+11+\mathrm{.....}+\left(3k+2\right)=\frac{k\left(3k+7\right)}{2}$ is True

We need to prove it is True for $n=k+1$

Left hand side = $5+8+11+\mathrm{....}+\left(3k+2\right)+\left(3\left(k+1\right)+2\right)$

  $\Rightarrow$ Left hand side = $\frac{k\left(3k+7\right)}{2}+\left(3k+3+2\right)$

  $\Rightarrow$ Left hand side = $\frac{k\left(3k+7\right)}{2}+3k+5$

  $\Rightarrow$ Left hand side = $\frac{k\left(3k+7\right)+2\left(3k+5\right)}{2}$

  $\Rightarrow$ Left hand side = $\frac{3k^2+7k+6k+10}{2}$

  $\Rightarrow$ Left hand side = $\frac{3k^2+13k+10}{2}$

Right hand side = $\frac{\left(k+1\right)\left(3\left(k+1\right)+7\right)}{2}$

  $\Rightarrow$ Right hand side = $\frac{\left(k+1\right)\left(3k+3+7\right)}{2}$

  $\Rightarrow$ Right hand side = $\frac{\left(k+1\right)\left(3k+10\right)}{2}$

  $\Rightarrow$ Right hand side = $\frac{3k^2+10k+3k+10}{2}$

  $\Rightarrow$ Right hand side = $\frac{3k^2+13k+10}{2}$

Left hand side = Right hand side

And hence the proof by Mathematical induction.

is True for all natural numbers $n$ .

Given information:

Use Mathematical induction to prove the above mentioned formula.

Concept used:

1.By using Mathematical induction technique we need to prove that this formula holds True for all natural numbers. The first step is to prove True for $n=1$ .

2.Second step by using induction hypothesis of mathematical induction we assume for $n=k$ given formula is True.

3.Third step is we need to prove that above formula is True for $n=k+1$ by using formula is True for $n=k$ .

Proof:

First we need to prove the above formula is True for $n=1$ .

Given formula is

  $1+4+7+\mathrm{.....}+3n-2=\frac{n\left(3n-1\right)}{2}$

Put $n=1$

We get

Left hand side = 1

Right hand side = $\frac{n\left(3n-1\right)}{2}=\frac{1\left(3\left(1\right)-1\right)}{2}=\frac{2}{2}=1$

So Left hand side = Right hand side

It True for $n=1$ .

Let us assume it is True for $n=k$

So

  $1+4+7+\mathrm{.....}+3k-2=\frac{k\left(3k-1\right)}{2}$ is True

We need to prove it is True for $n=k+1$

Left hand side = $1+4+7+\mathrm{.....}+3k-2+3\left(k+1\right)-2$

  $\Rightarrow$ Left hand side = $\frac{k\left(3k-1\right)}{2}+3k+3-2$

  $\Rightarrow$ Left hand side = $\frac{k\left(3k-1\right)}{2}+3k+1$

  $\Rightarrow$ Left hand side = $\frac{3k^2-k+2\left(3k+1\right)}{2}$

  $\Rightarrow$ Left hand side = $\frac{3k^2-k+6k+2}{2}$

  $\Rightarrow$ Left hand side = $\frac{3k^2+5k+2}{2}$

Right hand side = $\frac{\left(k+1\right)\left(3\left(k+1\right)-1\right)}{2}$

  $\Rightarrow$ Right hand side = $\frac{\left(k+1\right)\left(3k+3-1\right)}{2}$

  $\Rightarrow$ Right hand side = $\frac{\left(k+1\right)\left(3k+2\right)}{2}$

  $\Rightarrow$ Right hand side = $\frac{3k^2+2k+3k+2}{2}$

  $\Rightarrow$ Right hand side = $\frac{3k^2+5k+2}{2}$

Left hand side = Right hand side

And hence the proof by Mathematical induction.