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Question Number 54646 by gunawan last updated on 08/Feb/19

Such That

a ._{n + 1} C_{r} = \frac{(n + 1) ._{n} C_{r}}{(n - r + 1)}

b ._{n} C_{0} +_{n} C_{2} +_{n} C_{4 \dots} =_{n} C_{1} +_{n} C_{3} +_{n} C_{5 \dots} = 2^{n - 1}

Answered by Kunal12588 last updated on 08/Feb/19

L H S =^{n + 1} C_{r} = \frac{(n + 1)!}{r! (n + 1 - r)!}

= \frac{(n + 1) n!}{r! (n + 1 - r) (n - r)!} = \frac{n + 1}{n - r + 1} \times \frac{n!}{r! (n - r)!}

= \frac{(n + 1)^{n} C_{r}}{(n + 1 - r)} = R H S p r o v e d

Commented by gunawan last updated on 08/Feb/19

thank you Sir

Answered by Kunal12588 last updated on 10/Feb/19

b . f r o m b i n i m i a l e x p a n s i o n

0^{n} = {(1 - 1)}^{n}

\Rightarrow 0 =^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - \dots . . + {(- 1)}^{n} \cdot^{n} C_{n}

\Rightarrow^{n} C_{0} +^{n} C_{2} +^{n} C_{4} + \dots . =^{n} C_{1} +^{n} C_{3} +^{n} C_{5} + \dots . e q (1)

L H S o f e q (1)

^{n} C_{0} +^{n} C_{2} +^{n} C_{4} + \dots .

=^{n - 1} C_{0} +^{n - 1} C_{1} +^{n - 1} C_{2} +^{n - 1} C_{3} +^{n - 1} C_{4} + \dots

= 2^{n - 1} p r o v e d

I u s e d t h e i d e n t i t i e s (1) a n d (3) f r o m

q u e s t i o n 54740 s e e c a r e f u l l y