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Prove the following by using the principle of mathematical induction for all n ∈ N: (2n +7) < (n + 3)2
CBSE | Class 11 | Exercise 4.1 | Maths | NCERT | Principle of Mathematical Induction
Prove the following by using the principle of mathematical induction for all n ∈ N: (2n +7) < (n + 3)2

We can compose the given assertion as

    \[P\left( n \right):\text{ }\left( 2n\text{ }+7 \right)\text{ }<\text{ }\left( n\text{ }+\text{ }3 \right)2\]

In the event that

    \[n\text{ }=\text{ }1\]

we get

2.1+7=9<{{(1+3)}^{2}}=16

Which is valid.

Think about

    \[P\text{ }\left( k \right)\]

be valid for some certain whole number

    \[k\]

(2k+7)<{{(k+3)}^{2}}\text{ n}(1)

Presently let us demonstrate that

    \[P\text{ }\left( k\text{ }+\text{ }1 \right)\]

is valid.

Here

    \[\left\{ 2\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }+\text{ }7 \right\}\text{ }=\text{ }\left( 2k\text{ }+\text{ }7 \right)\text{ }+\text{ }2\]

We can compose it as

    \[=\text{ }\left\{ 2\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }+\text{ }7 \right\}\]

From condition (1) we get

(2k+7)+2<{{(k+3)}^{2}}+2

By growing the terms

2(k+1)+7<{{k}^{2}}+6k+9+2

On additional estimation

2(k+1)+7<{{k}^{2}}+6k+11

Here {{k}^{2}}+6k+11<{{k}^{2}}+8k+16

We can compose it as

    <span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno"> </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-062f6dd30e56ed8f7349bfda204ddd0d_l3.png" height="90" width="371" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & 2(k+1)+7<{{(k+4)}^{2}} \\ & 2(k+1)+7<(k+1)+{{3}^{2}} \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>

    \[P\text{ }\left( k\text{ }+\text{ }1 \right)\]

is valid at whatever point

    \[P\text{ }\left( k \right)\]

is valid.

Consequently, by the guideline of numerical enlistment, proclamation

    \[P\text{ }\left( n \right)\]

is valid for all normal numbers for example

    \[n.\]

i

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