Đáp án: Giải thích các bước giải: $\begin{array}{l}1 + 2.2 + {3.2^2} + \cdots + {100.2^{99}}\\ = (1 + 2 + {2^2} + \cdots + {2^{99}}) + (2 + {2^2} + \cdots + {2^{99}}) + \cdots + ({2^{98}} + {2^{99}}) + {2^{99}}\\ = {2^{100}} – 1 + 2({2^{99}} – 1) + {2^2}({2^{98}} – 1) + \cdots + {2^{98}}(2 – 1) + {2^{99}}\\ = {99.2^{100}} + {2^{99}} – (1 + 2 + {2^2} + \cdots {2^{98}})\\ = {99.2^{100}} + {2^{99}} – ({2^{99}} – 1)\\ = {99.2^{100}} + 1\end{array}$ Bình luận
1+2.2+3.22+⋯+100.299=(1+2+22+⋯+299)+(2+22+⋯+299)+⋯+(298+299)+299=2100−1+2(299−1)+22(298−1)+⋯+298(2−1)+299=99.2100+299−(1+2+22+⋯298)=99.2100+299−(299−1)=99.2100+1 Bình luận
Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
1 + 2.2 + {3.2^2} + \cdots + {100.2^{99}}\\
= (1 + 2 + {2^2} + \cdots + {2^{99}}) + (2 + {2^2} + \cdots + {2^{99}}) + \cdots + ({2^{98}} + {2^{99}}) + {2^{99}}\\
= {2^{100}} – 1 + 2({2^{99}} – 1) + {2^2}({2^{98}} – 1) + \cdots + {2^{98}}(2 – 1) + {2^{99}}\\
= {99.2^{100}} + {2^{99}} – (1 + 2 + {2^2} + \cdots {2^{98}})\\
= {99.2^{100}} + {2^{99}} – ({2^{99}} – 1)\\
= {99.2^{100}} + 1
\end{array}$
1+2.2+3.22+⋯+100.299=(1+2+22+⋯+299)+(2+22+⋯+299)+⋯+(298+299)+299=2100−1+2(299−1)+22(298−1)+⋯+298(2−1)+299=99.2100+299−(1+2+22+⋯298)=99.2100+299−(299−1)=99.2100+1