如何计算下面的公式

Question

考虑，

1*2^1 + 2*2^2 + 3*2^3 + 4*2^4 + ... d * 2^d 

= sum(r * 2^r, r from 1 to d) 

我们如何可以推断以下解决方案？

= 2 (d-1) * 2^d + 2 

谢谢

Answer 1

通过induction上d：

基本情况

d = 1 
sum(r * 2^r, r from 1 to 1) = 1 * 2^1 = 1 * 2 = 2 
2 * (1 - 1) * 2^1 + 2 = 2 * 0 * 2 + 2 = 0 + 2 = 2 

感应案例

我们假定归纳假设是正确的d这样的：

sum(r * 2^r, r from 1 to d + 1) = 
sum(r * 2^r, r from 1 to d) + [(d + 1) * 2^(d + 1)] = 
2 * (d-1) * 2^d + 2 + [(d + 1) * 2^(d + 1)] = 
(d - 1) * 2^(d + 1) + 2 + d * 2^(d + 1) + 2^(d + 1) = 
d * 2^(d + 1) - 2^(d + 1) + 2 + d * 2^(d + 1) + 2^(d + 1) = 
d * 2^(d + 1) + 2 + d * 2^(d + 1) = 
2 * d * 2^(d + 1) + 2 (result 1) 

和现在评估公式为d + 1

2 (d-1) * 2^d + 2 = (substituting d + 1 for d) 
2 * (d + 1 - 1) * 2^(d + 1) + 2 = 
2 * d * 2^(d + 1) + 2 (result 2) 

从而

2 * d * 2^(d + 1) + 2 (result 1) = 2 * d * 2^(d + 1) + 2 (result 2) 

QED

Answer 2

我认为你可以证明是用归纳显示此：

http://en.wikipedia.org/wiki/Proof_by_induction