1+3^2+3^3+...+3^200
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PT
1
NT
11 tháng 2 2017
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+....+\frac{1}{200}\left(1+2+...+200\right)\\ \Rightarrow E=1+\frac{1}{2}\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{200}\frac{\left(1+200\right).200}{2}\\ \Rightarrow E=1+\frac{1+2}{2}+\frac{1+3}{2}+....+\frac{1+200}{2}\\ \Rightarrow E=1+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\\ \Rightarrow E=\frac{2+3+4+....+201}{2}=\frac{\left(201+2\right).200:2}{2}=10150\)
Chúc bạn học tốt !!!
PT
0
VD
1
PT
20 tháng 2 2017
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(E=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{200}.\frac{\left(1+200\right).200}{2}\)
\(E=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{1+200}{2}\)
\(E=1+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)
\(E=\frac{2+3+4+...+201}{2}=\frac{\left(201+2\right).200:2}{2}\)
\(E=10150\)
SC
0
NT
0
10 tháng 9 2023
\(E=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{200}\cdot\dfrac{200\cdot201}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{201}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+...+\dfrac{201}{2}\)
\(=\dfrac{\left(201-2+1\right)\cdot\left(201+2\right)}{4}=\dfrac{200\cdot203}{4}=50\cdot203=10150\)
HQ
0
30 tháng 6 2018
Áp dụng công thức \(1+2+...+n=\frac{n\left(n+1\right)}{2}\)ta có:
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}=\frac{\frac{201.202}{2}-1}{2}=10150\)
A = 1 + 32 + 33 + 3200
3A = 3 + 33 + 34 + 35 + ... + 3201
3A - A = 3201 - 4
A = \(\frac{3^{201}-4}{2}\)
Hk tốt
\(A=1+3^2+3^3+...+3^{200}\)
\(A=1+\left(3^2+3^3+...+3^{200}\right)\)
Đặt \(B=3^2+3^3+...+3^{200}\)
\(3B=3^3+3^4+...+3^{201}\)
\(2B=3^{201}-3^2\Rightarrow B=\frac{3^{201}-9}{2}\)
Thế vào ta có: \(A=1+\left(3^2+3^3+...+3^{200}\right)=1+\frac{3^{201}-9}{2}\)