Show that- 1 - 1- 1 - 2 - 1- 1 - 2 - 3 - - - 1- 1 - 2 - 3 - -n - 2n- n - 1

1 + 1( 1 + 2) + 1( 1 + 2 + 3) + ... + 1( 1 + 2 + 3 + ...n) = 2n( n + 1) Let r^th term be Tr = 2r(r+1) = 2 [ 1r 1r+1 ] where r = 1 , 2 , 3 . ...

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Permutation under Restriction

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Question

Show that:
$1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + . . . + \frac{1}{(1 + 2 + 3 + . . . n)} = \frac{2 n}{(n + 1)}$

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

1 - 2 n + \frac{2 n (2 n - 1)}{2!} - \frac{2 n (2 n - 1) (2 n - 1)}{3!} + . . . + (- 1)^{n - 1} \frac{2 n (n - 1) . . . (n + 2)}{(n - 1)}

= (- 1)^{n + 1} \frac{(2 n)}{2 (n!)^{2}},

1^{2} + 3^{2} + 5^{2} + . . . + {(2 n - 1)}^{2} = \frac{n (2 n - 1) (2 n + 1)}{3} .

1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + . . . . . \frac{1}{1 + 2 + 3 + . . . . n} = \frac{2 n}{n + 1}

\frac{1^{2}}{(1) (3)} + \frac{2^{2}}{(3) (5)} + . . . . . + \frac{n^{2}}{(2 n - 1) (2 n + 1)} = \frac{n (n + 1)}{2 (2 n + 1)}

\frac{1}{(1) (3)} + \frac{1}{(3) (5)} + . . . . . + \frac{1}{(2 n - 1) (2 n + 1)} = \frac{1}{2 n + 1}

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