Question

Prove by method of induction, for all n€N 1) 8+17+26+.................+(9n-1)=n/2(9n+7)

Answer 1

$\text{Let P(n) denote the statement:}$

$8+17+26+.......+(9n-1)=\displaystyle\frac{n(9n+7)}{2}$

$\text{P(1):}$

$\text{L.H.S= 8}$

$\text{R.H.S=}\displaystyle\frac{1(9(1)+7)}{2}=\frac{16}{2}=8$

$\text{L.H.S=R.H.S}$

$\therefore\text{P(1) is true}$

$\text{Assume the result is true for n=k}$

$\text{That is, P(k) is true}$

$\text{That is}\;8+17+26+.......+(9k-1)=\displaystyle\frac{k(9k+7)}{2}\;\text{is true}$.......(1)

$\text{To prove: P(k+1) is true}$

$\textbf{That is to prove:}\;8+17+26+.......+(9(k+1)-1)=\displaystyle\frac{(k+1)(9k+16)}{2}$

$\text{Consider,}$

$8+17+26+.......+(9k-1)+(9(k+1)-1)$

$=\displaystyle\frac{k(9k+7)}{2}+(9k+8)$

$=\displaystyle\frac{k(9k+7)+18k+16}{2}$

$=\displaystyle\frac{9k^2+7k+18k+16}{2}$

$=\displaystyle\frac{9k^2+25k+16}{2}$

$=\displaystyle\frac{9k^2+9k+16k+16}{2}$

$=\displaystyle\frac{9k(k+1)+16(k+1)}{2}$

$=\displaystyle\frac{(k+1)(9k+16)}{2}$

$\therefore\text{P(k+1) is true}$

$\text{Hence, by mathematical induction P(n)}$

$\text{is true for all natural numbers}$