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Evaluate \binom{4}{0}+2\binom{4}{1}+3\binom{4}{2}+4\binom{4}{3}+5\binom{4}{4}

Option: 1

40


Option: 2

36


Option: 3

48


Option: 4

64


Answers (1)

best_answer

As we learnt

\dpi{120} C_{1}+2.C_{2}+3.C_{3}+---+n.C_{n}= \sum_{r=0}^{n}r.^{n}C_{r}=n\cdot 2^{n-1}

and

C_{0}+C_{1}+C_{2}+C_{3}+----+C_n= \sum_{r=0}^{n} (^{n} C_{r}) = 2^{n}

Now,

Given expression can be written as 

\sum_{r=0}^{4} (r+1) ^{4} C_{r}

=\sum_{r=0}^{4} (r) .^{4} C_{r} + \sum_{r=0}^{4} (^{4} C_{r})

= 4.2^3 + 2^4 = 32 + 16 = 48

 

Posted by

Suraj Bhandari

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