Question

Prove that : $(2n+7)<(n+3{)}^2$

Answer 1

Let's,

P(n) = $(2n+7)<(n+3{)}^2$

【For n = 1】

So,

L.H.S,

2n+7

= (2 × 1) + 7

=9

R.H.S,

$(n+3{)}^3$

$=(1+3{)}^3$

$=(4{)}^3$

$=16$

[Note:- the value of 'n' can be anything]

Hence, We Can Say, $(2n+7)<(n+3{)}^3$

Another Process

Let's, P(k) is True.

(2k+7)<$(n+3{)}^2$ .......(i)

L.H.S = (2(k+1)+7)

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

L.H.S

(2(k+1)+7)

=2k+2+7

=(2k+7)+2

[Using (i)]

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

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

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

$<k^2+6k+11$

R.H.S

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

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

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

$=k^2+8k+16$

Now, Compare Both Sides,

6k < 8k

11 < 16

So We Can Say,

$(2n+7)<(n+3{)}^2$