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Using the principle of mathematical induction, prove each of the following for all n ϵ N:

\(\frac1{2\times5}\) + \(\frac1{(5\times8)}\)+......+ \(\frac{1}{(3n - 1) \times(3n+2)}\) = \(\frac{n}{(6n+4)}\)

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To Prove:

\(\frac1{2\times5}\) + \(\frac1{(5\times8)}\)+......+ \(\frac{1}{(3n - 1) \times(3n+2)}\) = \(\frac{n}{(6n+4)}\)

For n = 1 

LHS = \(\frac1{2\times5}\)\(\frac1{10}\)

RHS = \(\frac{1\times1}{(6+4)}\) = \(\frac1{10}\)

Hence, LHS = RHS 

P(n) is true for n = 1 

Assume P(k) is true

\(\frac1{2\times5}\) + \(\frac1{(5\times8)}\)+......+ \(\frac{1}{(3k - 1) \times(3k+2)}\) = \(\frac{k}{(6k+4)}\).....(1)

We will prove that P(k + 1) is true

RHS = \(\frac{k+1}{(6(k+1)+4)}\) = \(\frac{k+1}{(6k+10)}\)

LHS = \(\frac1{2\times5}\) + \(\frac1{(5\times8)}\)+......+ \(\frac{1}{(3k - 1) \times(3k+2)}\) + \(\frac{1}{(3(k + 1)-1) \times(3(k+1)+2)}\)

[Writing the Last second term]

\(\frac1{2\times5}\) + \(\frac1{(5\times8)}\)+......+ \(\frac{1}{(3k - 1) \times(3k+2)}\) + \(\frac{1}{(3(k + 1)-1) \times(3(k+1)+2)}\)

(Taking LCM and simplifying)

\(\frac{k+1}{(6k+10)}\)

= RHS 

Therefore, \(\frac1{2\times5}\) + \(\frac1{(5\times8)}\)+......+ \(\frac{1}{(3k - 1) \times(3k+2)}\) + \(\frac{1}{(3(k + 1)-1) \times(3(k+1)+2)}\) = \(\frac{k+1}{(6k+10)}\)

LHS = RHS 

Therefore, P (k + 1) is true whenever P(k) is true 

By the principle of mathematical induction, P(n) is true for 

Where n is a natural number 

Put k = n - 1

\(\frac1{2\times5}\) + \(\frac1{(5\times8)}\)+......+ \(\frac{1}{(3n - 1) \times(3n+2)}\) = \(\frac{n}{(6n+4)}\)

Hence proved.

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