Question

By using the principal of mathematical induction prove that
$\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{ 3.4.5} + ... + \frac{1}{n(n + 1)(n + 2)} = \frac{n(n + 3)}{4(n + 1)(n + 2)}$
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Answer 1

Hope this answer helps you !

Answer 2

Basic question of induction

Step1):-First check out whether LHS is satisfying with RHS or notputting n=1 we getLHS=

$\frac{1}{1.2.3}$

now,in RHS ,

putting n=1 we getRHS=

$\frac{1 \times 4}{4 \times2 \times 3} = \frac{1}{1.2.3}$

from this we can sayLHS is satisfying RHS,,i.e =>

LHS=RHS

Step2):-Now let us check whether the LHS and RHS is satisfying at n=k+1 or notw

LHS at n=k+1,

$\frac{1}{1.2.3} \times \frac{1}{2.3.4} + ..... + \frac{1}{k(k + 1)(k + 2)} + \frac{1}{(k + 1)(k + 2)(k + 3)}$upto 1/k(k+1)(k+1) terms, sum=

{k(k+3)}/{4(k+1)(k+2)}

putting this value we getLHS=

$\frac{k(k + 3)}{4(k + 1)(k + 2)} + \frac{1}{(k + 1)(k + 2)(k + 3)} \\ lhs = \frac{k( {k + 3)}^{2} + 4}{4(k + 1)(k + 2)(k + 3)} = \frac{ {k}^{3} + 9k + 6 {k}^{2} + 4 }{4(k + 1)(k + 2)(k + 3)} \\ factorizing \: the \: numerator \\ lhs = \frac{(k + 1)( {k}^{2} + 5k + 4 )}{4(k + 1)(k + 2)(k + 3)} = \frac{(k + 1)(k + 4)}{4(k + 2)(k + 3)}$to take out RHS at , at n=k+1 just put k+1 in place of n

$\frac{(k + 1)(k + 4)}{4(k + 2)(k + 3)}$

This again shows LHS=RHShence for any value of n,,LHS will always be equal to RHS