# Using the principle of mathematical induction, prove each of the following for all n ϵ N: 1/2 x 5 + 1/(5 x 8)+......+ 1/(3n - 1) x (3n +2)

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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.