Question

For every n $\in$ $N$,
the value of $\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}$is an

• A. even natural number
• B. odd natural number
• C. natural number
• D. irrational number

Answer 1

The correct option is C natural number

$\mathrm{Let}P\left(n\right)=\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}\mathrm{Then}P\left(1\right)=\frac{1}{5}+\frac{1}{3}+\frac{7}{15}=1,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{natural}\mathrm{number}.P\left(2\right)=\frac{2^5}{5}+\frac{2^3}{3}+\frac{7\times 2}{15}=\frac{32}{5}+\frac{8}{3}+\frac{14}{15}=10,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{number}.\mathrm{Thus},P\left(1\right)\mathrm{and}P\left(2\right)\mathrm{both}\mathrm{are}\mathrm{natural}\mathrm{number}.\mathrm{Let}\mathrm{us}\mathrm{assume}\mathrm{that}P\left(k\right)\mathrm{is}a\mathrm{natural}\mathrm{number}\mathrm{for}k\in N,i.e.,\frac{k^5}{5}+\frac{k^3}{3}+\frac{7k}{15}=m\mathrm{for}\mathrm{some}m\in N...\left(1\right)\mathrm{Now}P\left(k+1\right)=\frac{{\left(k+1\right)}^5}{5}+\frac{{\left(k+1\right)}^3}{3}+\frac{7\left(k+1\right)}{15}=\frac{k^5+5k^4+10k^3+10k^2+5k+1}{5}+\frac{k^3+3k^2+3k+1}{3}+\frac{7k}{15}+\frac{7}{15}=\left(\frac{k^5}{5}+\frac{k^3}{3}+\frac{7k}{15}\right)+\left(k^4+2k^3+3k^2+2k\right)+\left(\frac{1}{5}+\frac{1}{3}+\frac{7}{15}\right)=m+\left(k^4+2k^3+3k^2+2k\right)+P\left(1\right)\left(\mathrm{Using}\left(1\right)\right]\mathrm{which}\mathrm{is}\mathrm{again}a\mathrm{natural}\mathrm{number}\mathrm{as}k\in N\mathrm{So},P\left(k+1\right)\mathrm{is}\mathrm{true}\mathrm{whenever}P\left(k\right)\mathrm{is}\mathrm{true}.\mathrm{Hence},\mathrm{by}\mathrm{the}\mathrm{principle}\mathrm{of}\mathrm{mathematical}\mathrm{induction},\mathrm{the}\mathrm{value}\mathrm{of}P\left(n\right)\mathrm{is}a\mathrm{natural}\mathrm{number}\mathrm{for}\mathrm{all}n\in N.$