Using the Mathematical induction- show that for any number n - 2- 11 - 2 - 11 - 2 - 3 - 11 - 2 - 3 - 4 - - -11 - 2 - 3 - - - n - n - 1n - 1

[ 11 + 2 + 11 + 2 + 3 + 11 + 2 + 3 + 4 + .... + 11 + 2 + 3 + .... + n = n 1n + 1; 11 + 2 + 11 + 2 + 3 + 11 + 2 + 3 + ...

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Standard XII

Mathematics

Proof by mathematical induction

Using the Mat...

Question

Using the Mathematical induction, show that for any number $n \geq 2,$
$\frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + . . . . . + \frac{1}{1 + 2 + 3 + . . . + n} = \frac{n - 1}{n + 1}$

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Solution

$\begin{matrix} \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + . . . . + \frac{1}{1 + 2 + 3 + . . . . + n} = \frac{n - 1}{n + 1} \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + . . . . T_{n} = \frac{1}{1 + 2 + 3 + . . . . . (n + 1)} = \frac{1}{\frac{(n + 1) (n + 2)}{2}} = \frac{2}{(n + 1) (n + 2)} = 2 [\frac{1}{(n + 1)} - \frac{1}{(n + 2)}] T_{1} = 2 [\frac{1}{2} - \frac{1}{3}], T_{2} = 2 [\frac{1}{3} - \frac{1}{4}], . . . . T_{n} = 2 [\frac{1}{(n + 1)} - \frac{1}{(n + 2)}] S_{n} = 2 [\frac{1}{2} - \frac{1}{(n + 1)}] = \frac{n - 1}{n + 1} \end{matrix}$

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Using the Mathematical induction, show that for any number n≥2,11+2+11+2+3+11+2+3+4+.....+11+2+3+...+n=n−1n+1

11+2+11+2+3+11+2+3+4+....+11+2+3+....+n=n−1n+111+2+11+2+3+11+2+3+4+....Tn=11+2+3+.....(n+1)=1(n+1)(n+2)2=2(n+1)(n+2)=2[1(n+1)−1(n+2)]T1=2[12−13],T2=2[13−14],....Tn=2[1(n+1)−1(n+2)]Sn=2[12−1(n+1)]=n−1n+1

Using the Mathematical induction, show that for any number $n \geq 2,$
$\frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + . . . . . + \frac{1}{1 + 2 + 3 + . . . + n} = \frac{n - 1}{n + 1}$