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The value of \sum_{r=0}^{n}(r+1)^{n}C_{r} equals

 

Option: 1

(n+2)2^{n-1}


Option: 2

(n+1)2^{n-1}


Option: 3

(n+2)2^{n}


Option: 4

(n+4)2^{n-1}


Answers (1)

best_answer

As we have learned

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= 2^{n}

 

 

Now,

We have  , 

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

(^nC_1 + 2 ^nC_2+ ....+n ^nC_n)+ (^nC_0+ ^nC_1+....+^nC_n)

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

Posted by

vinayak

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