Question

ahoy!!
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prove this by PMI.

(2n +7) < (n + 3)^2

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:) thank you !

Answer 1

Heya!!

★Answer★
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Given→ (2n + 7) < (n + 3)^2

Now,,

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

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◆Another Process Which You Want :-

Let's, P(k) is True.

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


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

R.H.S = ((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

((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

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#Hope Helped (:

Answer 2

Answer:

Heya!! buddy this is ur answer!!