Prove the following by using the principle of mathematical induction for all $n in N: Question: Prove the following by using the principle of mathematical induction for all n ∈ N$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{(6 n+4)}$Solution: Let the given statement be P(n), i.e.,$\mathrm{P}(n): \frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{(6 n+4)}$For n = 1, we have$\mathrm{P}(1)=\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}=\frac{1}{10}$, which is true. Let$\mathrm{P}(k)$be true for some positive integer$k$, i.e.,$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 k-1)(3 k+2)}=\frac{k}{6 k+4}\ldots$(i) We shall now prove that$\mathrm{P}(k+1)$is true. Consider$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots \ldots+\frac{1}{(3 k-1)(3 k+2)}+\frac{1}{\{3(k+1)-1\}\{3(k+1)+2\}}=\frac{k}{6 k+4}+\frac{1}{(3 k+3-1)(3 k+3+2)}$[Using (i)]$=\frac{k}{6 k+4}+\frac{1}{(3 k+2)(3 k+5)}=\frac{k}{2(3 k+2)}+\frac{1}{(3 k+2)(3 k+5)}=\frac{1}{(3 k+2)}\left(\frac{k}{2}+\frac{1}{3 k+5}\right)=\frac{1}{(3 k+2)}\left(\frac{k(3 k+5)+2}{2(3 k+5)}\right)=\frac{1}{(3 k+2)}\left(\frac{3 k^{2}+5 k+2}{2(3 k+5)}\right)=\frac{1}{(3 k+2)}\left(\frac{(3 k+2)(k+1)}{2(3 k+5)}\right)=\frac{(k+1)}{6 k+10}=\frac{(k+1)}{6(k+1)+4}$Thus,$\mathrm{P}(k+1)$is true whenever$\mathrm{P}(k)$is true. Hence, by the principle of mathematical induction, statement$\mathrm{P}(n)$is true for all natural numbers i.e.,$n\$.