What is the sum of the series nCr . 2r?

[cgchayan]

what will be the sum of this series?
$$\Sigma$$ ⁿC_r . 2^r
where r = 0 to n

 

[l'Hôpital]

Hint:
$$\sum_{r=0}^n _nC_r \cdot 2^r = \sum_{r=0}^n _nC_r \cdot 2^r \cdot 1^{n-r}$$

 

[cgchayan]

thanx

 

[Sherard]

thanks for sharing information...

 

[blue_raver22]

The sum of the series nCr . 2r, where r = 0 to n, is equal to 2^n. This can be proven using the binomial theorem, where (1+2)^n is expanded to (nC0)(2^0) + (nC1)(2^1) + (nC2)(2^2) + ... + (nCn)(2^n). This is equivalent to the given series, where nCr is the coefficient of the term (2^r). Thus, the sum of the series is 2^n.