Question

Prove: $\left(2n+7\right)<{\left(n+3\right)}^2.$ for $n\in N$.

Answer 1

Let $p\left(n\right):(2n+7)<(n+3{)}^2$

For n=1

$L.H.S=(2.1+7)=2+7=9$

$R.H.S=(1+3{)}^2=4^2=16$

Since 9 < 16

L.H.S < R.H.S

$\therefore p\left(n\right)$ is true for n=1

Assume p(k) is true

$(2k+7)<(k+3{)}^2$ ______ (1)

we will prove that p(k+1) is true

$R.H.S=\left(\right(k+1)+3{)}^2$

$L.H.S=\left(2\right(k+1)+7)$

L.H.S

$\left[2\right(k+1)+7]$

$=2k+2+7$

$=(2k+7)+2$

Using (1) : $(2k+7)<(k+3{)}^2$

$<(k+3{)}^2+2$

$<k^2+9+6k+2$

$<k^2+6k+11$

$<k^2+6k+11+(2k+5)$

$<k^2+8k+16$

R.H.S

$\left[\right(k+1)+3{]}^2$

$=(k+4{)}^2$

$=k^2+4^2+\mathrm{2.4.}k$

$=k^2+8k+16$

L.H.S < R.H.S

$\therefore p(k+1)$ is true whenever p(k) is true.