Question

1/2×5+1/5×8+1/8×11+.....+1/(3n-1)(3n+1)=n/6n+4 prove by mathematical induction
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Answer 1

A small mistake in the question. Actually we have to prove,

$\displaystyle P(n)\ :\ \sum_{i=1}^{n}\dfrac{1}{(3i-1)(3i+2)}\ =\ \dfrac{n}{6n+4}$

$\cline{1-}$

Put n = 1.  Let's know whether P(1) is correct.

$\textsf{LHS\ =\dfrac{1}{2\cdot 5}=\dfrac{1}{10}=\dfrac{1}{6+4}= \ RHS}$

Hence P(1) is true. So we can assume that P(k) is true.

$\textsf{Assume\ \ P(k)\ :\ \displaystyle \sum_{i=1}^{k}\dfrac{1}{(3i-1)(3i+2)}\ =\ \dfrac{k}{6k+4} \ \ is true.}$

Now we are assuming P(k+1).

$\begin{aligned}&\textsf{LHS}\\ \\ \Longrightarrow\ \ &\sum_{i=1}^{k+1}\dfrac{1}{(3i-1)(3i+2)}\\ \\ \Longrightarrow\ \ &\sum_{i=1}^{k}\dfrac{1}{(3i-1)(3i+2)}+\dfrac{1}{(3(k+1)-1)(3(k+1)+2)}\\ \\ \Longrightarrow\ \ &\dfrac{k}{6k+4}+\dfrac{1}{(3k+3-1)(3k+3+2)}\\ \\ \Longrightarrow\ \ &\dfrac{k}{2(3k+2)}+\dfrac{1}{(3k+2)(3k+5)}\\ \\ \Longrightarrow\ \ &\dfrac{1}{3k+2}\left(\dfrac{k}{2}+\dfrac{1}{3k+5}\right)\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\dfrac{1}{3k+2}\left(\dfrac{k(3k+5)+2}{2(3k+5)}\right)\\ \\ \Longrightarrow\ \ &\dfrac{1}{3k+2}\left(\dfrac{3k^2+5k+2}{6k+10}\right)\\ \\ \Longrightarrow\ \ &\dfrac{1}{3k+2}\left(\dfrac{3k^2+3k+2k+2}{6k+6+4}\right)\\ \\ \Longrightarrow\ \ &\dfrac{1}{3k+2}\left(\dfrac{3k(k+1)+2(k+1)}{6(k+1)+4}\right)\\ \\ \Longrightarrow\ \ &\dfrac{1}{3k+2}\left(\dfrac{(3k+2)(k+1)}{6(k+1)+4}\right)\\ \\ \Longrightarrow\ \ &\dfrac{k+1}{6(k+1)+4}\end{aligned}$

$\begin{aligned}\\ \\ \Longrightarrow\ \ &\dfrac{n}{6n+4}\\ \\ \Longrightarrow\ \ &\textsf{RHS}\end{aligned}$

Hence Proved!