3-C-1-4-C-2-5-C-3-49-C-47-where-n-C-r-n-r-n-r-

Question Number 25122 by math solver last updated on 04/Dec/17

3_{C_{1}} + 4_{C_{2}} + 5_{C_{3}} + \dots \dots \dots . . + 49_{C_{47}} = ?

w h e r e n_{C_{r}} = \frac{n!}{r! \times (n - r)!} .

Commented by moxhix last updated on 04/Dec/17

d e f i n e n_{C_{r}} =:_{n} C_{r}

_{3} C_{1} +_{4} C_{2} +_{5} C_{3} + \dots +_{49} C_{47}

=_{3} C_{2} +_{4} C_{2} +_{5} C_{2} + \dots +_{49} C_{2}

= \frac{3 \times 2}{2 \times 1} + \frac{4 \times 3}{2 \times 1} + \frac{5 \times 4}{2 \times 1} + \dots + \frac{49 \times 48}{2 \times 1}

= \frac{1}{2} \underset{k = 1}{\sum^{47}} (k + 1) (k + 2)

= \frac{1}{2} \underset{k = 1}{\sum^{47}} (k^{2} + 3 k + 2)

= \frac{1}{2} (\frac{1}{6} 47 \times 48 \times 95 + \frac{3}{2} 47 \times 48 + 2 \times 47)

= 47 \times 4 \times 95 + 3 \times 47 \times 12 + 1 \times 47)

= 47 (4 \times 95 + 3 \times 12 + 1)

= 47 \times 417

= 19599

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