Question

Let f(n) = $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}$ such that P(n)f(n + 2) = P(n)f(n) + q(n) then match the following List I with List II

Answer 1

If $f\left(x\right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{n}$
$f(n+2)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}$
$f(n+2)-f\left(n\right)=\frac{2n+3}{(n+1)(n+2)}$
(n + 1)(n + 2) f(n + 2) = (n + 1)(n + 2) f(n) + 2n + 3
$p\left(n\right)=n^2+3n+2,q\left(n\right)=2n+3$
(P) $\sum _{n=1}^m(n+3)=\sum _{n=1}^{m}n+3\sum _{n=1}^{m}1=\frac{m(m+1)}{2}+3m=\frac{m(m+1-6)}{2}=\frac{m(m+7)}{2}$
(Q) $\sum _{n=1}^{m}n=\frac{m(m+1)}{2}$
(R) $\sum _{n=1}^m\frac{p\left(n\right)+q^2\left(n\right)-11}{n}=\sum _{n=1}^m\frac{n^2+3n+2+4n^2+9+12n-11}{n}=\sum _{n=1}^m\frac{5n^2+15n}{n}$
$=\sum _{n=1}^{m}5n+15=5\sum _{n=1}^{m}n+15\sum _{n=1}^{m}1=\frac{5m(m+1)}{2}+15m=\frac{5m(m+1+6)}{2}=\frac{5m(m+7)}{2}$
(S) $\sum _{n=1}^m\frac{q^2\left(n\right)-p\left(n\right)-7}{n}=\sum _{n=1}^{m}3n+9=3\sum _{n=1}^{m}n+9\sum _{n=1}^m=\frac{3m(m+1)}{2}+9m=\frac{3m(m+1+6)}{2}=\frac{3m(m+7)}{2}$