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., $n$.

 

 

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