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  • Show that n^5/5+n^3/3+7n/15 is a natural number for all n ϵ N.

Show that  \frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{15} is a natural number for all n ϵ N.

Answers (1)

Given data:

P(n) = \frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{15}

Now, we’ll substitute different values for n,

\begin{aligned} \text { At } n=1 & \\ P(1)=\frac{1^{5}}{5}+\frac{1^{8}}{3}+\frac{7(1)}{15} \\ &=\frac{1}{5}+\frac{1}{3}+\frac{7}{15} \end{aligned}

\\=(3+5+7 )/ 15\\ =1 \: \: is\: \: true.\\\\\ At \: \: $n=2$\\\\ $$ P(2)=\frac{2^{5}}{5} +\frac{2^{3}}{3}+\frac{7(2)}{15} \\\\ =\frac{32}{5}+\frac{8}{3}+\frac{14}{15} \\\\ =(96+40+14 )/ 15 =10 \: \: is\: \: true

At n=3

\\P(3)=\frac{3^{5}}{5} +\frac{3^{3}}{3}+\frac{7(3)}{15} \\\\ =\frac{243}{5}+\frac{27}{3}+\frac{21}{15} \\\\ =(729+135+21 )/ 15 \\\\=59is true.

Let us consider, 

P(k)=\frac{k^{5}}{5}+\frac{k^{2}}{3}+\frac{7(k)}{15} $$\: \: to\: \: be\: \: true.

Thus, at n=k+1

\\ P(k+1)=\frac{(k+1)^{5}}{5}+\frac{(k+1)^{3}}{3}+\frac{7(k+1)}{15} \\\\ =\frac{k^{5}+5 k^{4}+10 k^{2}+10 k^{2}+5 k+1}{5}+\frac{k^{3}+3 k^{2}+3 k+1}{3}+\frac{7(k+1)}{15} \\\\ =\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7(k)}{15}+k^{4}+2 k^{3}+3 k^{2}+2 k+1,\ is\ a\ natural\ number

Thus, n = k + 1 is true

Thus, by mathematical Induction,

For each natural no. n it is true that,

P(n) = \frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{15}

Posted by

infoexpert21

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