如何計算下面的公式

Question

考慮，

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

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

我們如何可以推斷以下解決方案？

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

謝謝

Answer 1

通過induction上d：

基本情況

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 

感應案例

我們假定歸納假設是正確的d這樣的：

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) 

和現在評估公式爲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) 

從而

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

QED

Answer 2

我認爲你可以證明是用歸納顯示此：

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