Chapter C, Problem 1E

To determine

To prove:

The formula for all natural number using the principle of induction.

Explanation of Solution

Given info.

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

If $\left(1\right)$ the statement is true for $n=1$ and

$\left(2\right)$ When a statement is true for a natural number $n=k$ then it will also be true for its successor $n=k+1$.

Then statement will be true for all natural number n. this is called principle of mathematical induction.

Proof:

Show that $P\left(1\right)$ is true

Left hand side $\left(L.H.S.\right)$

$P\left(1\right)=1$

R.H.S.

$\begin{array}{c}P\left(1\right)=\frac{1\times \left(1+1\right)}{2}\\ =\frac{1\times 2}{2}\\ =1\end{array}$

$L.H.S.=R.H.S.$

Thus, the statement is true for $n=1$

Now, follow induction step.

If $P\left(n\right)$ is true for $n=k$ then it will also be true for $n=k+1$

Let us assume that $P\left(k\right)$ is true condition that means

$P\left(k\right)=1+2+3+\mathrm{.........}+k=\frac{k\left(k+1\right)}{2}$

Then by taking above statement true, proceed and prove that its successor will also be true.

$P\left(k+1\right)=1+2+3+\mathrm{...........}+k+\left(k+1\right)=\frac{\left(k+1\right)\left(k+2\right)}{2}$

L.H.S.

$1+2+3+\mathrm{..........}+k+k+1$

As we know that

$\left(1+2+3+\mathrm{...........}+k\right)=\frac{k\left(k+1\right)}{2}$

Substitute the value

$\begin{array}{c}P\left(k+1\right)=\frac{k\left(k+1\right)}{2}+\left(k+1\right)\\ =\frac{k\left(k+1\right)+2\left(k+1\right)}{2}\end{array}$

Take $\left(k+1\right)$ as common factor

$P\left(k+1\right)=\frac{\left(k+1\right)\left(k+2\right)}{2}$

This is the value of R.H.S.

Now, we have full-filled both conditions of the principle of mathematical induction. The formula is therefore true for every natural number.