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Question Number 202251 by BaliramKumar last updated on 23/Dec/23

\frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \dots \dots \dots \dots . . + \frac{1}{n (n + 1) (n + 2)} = ?

Commented by BaliramKumar last updated on 23/Dec/23

\frac{1}{a (a + d) (a + 2 d)} + \frac{1}{(a + d) (a + 2 d) (a + 3 d)} + \dots \dots . + \frac{1}{{a + (n - 1) d} {a + nd} {a + (n + 1) d}} = ?

Commented by MATHEMATICSAM last updated on 24/Dec/23

T_{1} = \frac{1}{a (a + d) (a + 2 d)},

T_{2} = \frac{1}{(a + d) (a + 2 d) (a + 3 d)}, \dots \dots . .

T_{n} = \frac{1}{{a + (n - 1) d} {a + n d} {a + (n + 1) d}}

= \frac{1}{2 d} [\frac{{a + (n + 1) d} - {a + (n - 1) d}}{{a + (n - 1) d} {a + n d} {a + (n + 1) d}}]

= \frac{1}{2 d} [\frac{1}{{a + (n - 1) d} {a + n d}} - \frac{1}{{a + n d} {a + (n + 1) d}}]

T_{1} = \frac{1}{2 d} [\frac{1}{a (a + d)} - \frac{1}{(a + d) (a + 2 d)}]

T_{2} = \frac{1}{2 d} [\frac{1}{(a + d) (a + 2 d)} - \frac{1}{(a + 2 d) (a + 3 d)}]

T_{3} = \frac{1}{2 d} [\frac{1}{(a + 2 d) (a + 3 d)} - \frac{1}{(a + 3 d) (a + 4 d)}]

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T_{n - 1} = \frac{1}{2 d} [\frac{1}{{a + (n - 2) d} {a + (n - 1) d}} - \frac{1}{{a + (n - 1) d} {a + n d}}]

T_{n} = \frac{1}{2 d} [\frac{1}{{a + (n - 1) d} {a + n d}} - \frac{1}{{a + n d} {a + (n + 1) d}}]

Now T_{1} + T_{2} + T_{3} + \dots . + T_{n}

= \frac{1}{2 d} [\frac{1}{a (a + d)} - \frac{1}{{a + n d} {a + (n + 1) d}}]

Commented by BaliramKumar last updated on 24/Dec/23

Thanks Sir

we can also write in this form

= \frac{n}{2} [\frac{a + {a + (n + 1) d}}{a (a + d) {a + nd} {a + (n + 1) d}}]

Commented by MATHEMATICSAM last updated on 24/Dec/23

yes

Answered by MATHEMATICSAM last updated on 23/Dec/23

Let s = \frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \dots . + \frac{1}{n (n + 1) (n + 2)}

T_{1} = \frac{1}{1 \times 2 \times 3}, T_{2} = \frac{1}{2 \times 3 \times 4},

T_{3} = \frac{1}{3 \times 4 \times 5}, \dots \dots .

T_{n} = \frac{1}{n (n + 1) (n + 2)} = \frac{2}{n (n + 1) (n + 2)} \times \frac{1}{2}

= \frac{(n + 2) - n}{n (n + 1) (n + 2)} \times \frac{1}{2} = \frac{1}{2} [\frac{1}{n (n + 1)} - \frac{1}{(n + 1) (n + 2)}]

T_{1} = \frac{1}{2} [\frac{1}{1 \times 2} - \frac{1}{2 \times 3}]

T_{2} = \frac{1}{2} [\frac{1}{2 \times 3} - \frac{1}{3 \times 4}]

T_{3} = \frac{1}{2} [\frac{1}{3 \times 4} - \frac{1}{4 \times 5}]

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T_{n - 1} = \frac{1}{2} [\frac{1}{(n - 1) n} - \frac{1}{n (n + 1)}]

T_{n} = \frac{1}{2} [\frac{1}{n (n + 1)} - \frac{1}{(n + 1) (n + 2)}]

Now

s = T_{1} + T_{2} + T_{3} + \dots . + T_{n}

= \frac{1}{2} [\frac{1}{2} - \frac{1}{(n + 1) (n + 2)}]

= \frac{1}{2} [\frac{(n + 1) (n + 2) - 2}{2 (n + 1) (n + 2)}]

= \frac{n^{2} + 3 n + 2 - 2}{4 (n + 1) (n + 2)}

= \frac{n (n + 3)}{4 (n + 1) (n + 2)} (Ans)

Commented by BaliramKumar last updated on 23/Dec/23

Thanks Sir