UPSEE Solved Model Paper 6

We can solve this problem using statement of principle of mathematical induction. Statement of the Principle of Mathematical Induction: Let P (n) be a statement involving the natural number ‘n’ such that (i) P (1) is true i.e. P(n) is true for n =1 and (ii) P (m+1) is true, whenever P(m) is true i.e., P(m) is true ⇒ P(m+1) is true. Then, P (n) is true for all natural numbers n. By using these steps we can find the answer. Given problem is Summation of series type When n = 1, we have $\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}$ = $\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$ = $\frac{3+5+7}{15}$ = $\frac{15}{15}$ = 1 ∊ N We shall show that $\left(\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}\right)$ is a natural number for all n ∊ N. Let P (n) be the statement : $\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}$ is a natural number Basic step : For n = 1 $\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}$ = $\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$ = $\frac{3+5+7}{15}$ = $\frac{15}{15}$ = 1 ∊ N Thus P (1) is true. Induction step : Assume that P 9k) is true that is $\frac{k^5}{5}+\frac{k^3}{3}+\frac{7k}{15}$ = m ... (i) where m ∊ N now, to prove P (k + 1) is true $\left(\frac{(k+1)^5}{5}+\frac{(k+1)^3}{3}+\frac{7(k+1)}{15}\right)$ = = $\left(\frac{k^5}{5}+\frac{k^3}{3}+\frac{7k}{15}\right)$ + $\left(k^4+2k^3+2k^2+k+\frac{1}{5}\right)$ + $\left(k^2+k+\frac{1}{3}\right)+\frac{7}{15}$ = m + (Using (i)) = = $m+(k^4+2k^3+3k^2+2k)$ + 1 ∊ N [since m ∊ N and $(k^4+2k^3+3k^2+2k)$ + 1 ∊ N] Thus, P k + 1) is true whenever P (k) is true Hence, by mathematical induction P (n) is true for all n ∊ N i.e., $\left(\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}\right)$ is a natural number for all n ∊ N.