Tính tổng: $S=1+2*2+3*2^2+4*2^3+.....+100*2^{99}  $
bấm máy ra số khủng –  tran85295 02-02-15 08:24 PM
Xét f(x) = (x^101 - 1)/(x-1) = x^100 + x^99 + .... + x + 1 
=> f'(x) = (101x^100 - x^101 + 1)/(x-1)^2 = 100x^99 + 99x^98 + ... + 2x + 1 
=> f'(2) = A = (101 . 2^100 - 2^101 + 1) 
= 2^100 (101 - 2) + 1 = 99.2^100 + 1 

PS: Mình còn có thêm 1 cách nữa nè :-? 
Ta có 
A = 1 + 2.2 + 3.2^2 + .... + 100.2^99 
2A = 2 + 2.2^2 + >>> + 99.2^99 + 100.2^100 
=> A - 2A = 1 + 2 + 2^2 + ... + 2^99 - 100.2^100 
=> A = 100.2^100 - (1 + 2 + 2^2 + .... + 2^99) 
= 100.2^100 - 2^100 + 1 = 99.2^100 + 1
Đặt $A=2^0+2^1+2^2+....+2^{99}$
ta có $S=1.2^0+2.2^1+....+100.2^{99}$
$=(2^0+2^1+2^2+...+2^{99})+(0.2^0+1.2^1+2.2^2+....+99.2^{99})$
$=A+2.(1.2^0+2.2^1+3.2^2+....+99.2^{98})=A+2.(S-100.2^{99})$(*)
$\Rightarrow S=100.2^{100}-A$(*)
Lại có $2A=2^1+2^2+2^3+....+2^{100}=A-2^0+2^{100}$
tức là $A=2^{100}-2^0=2^{100}-1$
thế vào (*) ta được $S=100.2^{100}-2^{100}+1=99.2^{100}+1$

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