Ta có: $\sum_{i=0}^{100}x^i=(1+x)^{100}$$\Rightarrow \sum_{i=0}^{100}ix^{i-1}=100(1+x)^{99}$ (đạo hàm 2 vế)$\Leftrightarrow \sum_{i=1}^{100}ix^{i-1}=100(1+x)^{99}$
Ta có: $\sum_{i=0}^{100}x^i=
\dfrac{1
-x^{10
1}}{1-x}$$\Rightarrow \sum_{i=0}^{100}ix^{i-1}=
\dfrac{100
x^{101
}-101x^{100}+
1}{(x
-1)^
2}$ (đạo hàm 2 vế)$\Leftrightarrow \sum_{i=1}^{100}ix^{i-1}=
\dfrac{100
x^{101}-101x^{1
00}+
1}{(x
-1)^
2}$