Show that 1- 1 2 - 1 3 - - 1 n-1 - 1 1- 1 3- 4 5- 9 7- - n 2 2n-1-

We know that; 1 u _ 1 + 1 u _ 2 + 1 u _ 3 +....+ 1 u _ n+1 = 1 u _ 1 + u _ 1 ^ 2 u _ 1 + u _ 2 u _ 2 ^ 2 u _ 2 + u _ 3 ...... u _ n ^ 2 ...

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Equivalent Fractions

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Question

Show that
$1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n + 1} = \frac{1}{1 -} \frac{1}{3 -} \frac{4}{5 -} \frac{9}{7 -} \dots \frac{n^{2}}{2 n + 1}$ .

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\frac{2}{1 -} \frac{3}{5 -} \frac{8}{7 -} \dots \frac{n^{2} - 1}{2 n + 1} = \frac{n (n + 3)}{2}

$(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) \dots (1 + \frac{(2 n + 1)}{n^{2}}) = (n + 1)^{2}$

\frac{1}{1 -} \frac{4}{5 -} \frac{9}{13 -} \frac{64}{25 -} \dots \frac{{(n^{2} - 1)}^{2}}{n^{2} + {(n + 1)}^{2}} = \frac{(n + 1) (n + 2) (2 n + 3)}{6}

\frac{C_{0}}{1} - \frac{C_{1}}{4} + \frac{C_{2}}{9} - . . . + \frac{(- 1)^{n} C_{n}}{n^{2} + 2 n + 1}

= \frac{1}{n + 1} + \frac{1}{2 n + 2} + . . . + \frac{1}{n^{2} + 2 n + 1} .

\frac{1}{(1 + 2)} + \dots + \frac{1}{(1 + 2 + 3 \dots n)} = \frac{2 n}{(n + 1)}

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