(2n+7)>(n+3)² prove by mathematical induction
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Sia ? 6 years, 4 months ago
Let P (n) =(2n + 7) < (n + 3)2
For n = 1
{tex}P(1) = (2 \times 1 + 7) < {(1 + 3)^2} \Rightarrow 9 < 16{/tex}
{tex}\therefore {/tex} P ( 1) is true
Let P(n) be true for n = k
{tex}\therefore P(k) = (2k + 7) < {(k + 3)^2}{/tex} ....(1)
For n = k + 1
P (k + 1) = 2 (k + 1) + 7 < (k + 1 + 3)2
{tex} \Rightarrow {/tex} 2(k + 1) + 7 < (k + 4)2
From (1)
2k + 7 < (k + 3)2
Adding 2 on both sides
2k + 7 + 2 < (k + 3)2 + 2 {tex} \Rightarrow {/tex} 2(k + 1) + 7 < k2 + 9 + 6k + 2
{tex} \Rightarrow {/tex} 2(k + 1) + 7 < k2 +6k + 11 < k2 + 8k + 16
2(k + 1) + 7 <(k + 4)2
{tex}\therefore {/tex} P (k + 1) is true
Thus P(k) is true {tex} \Rightarrow {/tex} P(k + 1) is true
Hence by principle of mathematical induction, P (n) is true for all {tex}n \in N{/tex}.
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