a=1+4+4^2+4^3+4^200
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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) Ta có: A = 1 + 3 + 32 + ... + 3200
⇒ 3A = 3 + 32 + 33 + ... + 3200
⇒ 3A - A = (3 + 32 + 33 + ... + 3201) - (1 + 3 + 32 + ... + 3200)
⇒ 2A = 3201 - 1
⇒ A = 2A : A = \(\dfrac{3^{201}-1}{2}\)
Vậy A = \(\dfrac{3^{201}-1}{3}\)
a: =(-1)+(-1)+...+(-1)=-1011
b: =(-5)+(-5)+...+(-5)=-175
Ta đã biết \(\dfrac{1}{a\cdot a}< \dfrac{1}{\left(a+1\right)\left(a-1\right)}\) ( a ϵ Z )
⇒ \(Q=\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+\dfrac{1}{4\cdot4}+...+\dfrac{1}{200\cdot200}\) < \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
Ta có \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{199\cdot201}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{201}\right)=\dfrac{1}{2}\cdot\dfrac{200}{201}=\dfrac{100}{201}< \dfrac{100}{200}=\dfrac{1}{2}< \dfrac{3}{4}\)
Vậy Q < \(\dfrac{3}{4}\)
Bài 1:
A = 1 + 3 + 32 + ... + 3100
=> 3A = 3 + 32 + ... + 3101
=> 2A = 3101 - 1
=> A = \(\frac{3^{101}-1}{2}\)
B = 1 + 42 + 44 + ... + 4100
=> 8B = 42 + 44 + ... + 4102
=> 7B = 4102 - 1
=> B = \(\frac{4^{102}-1}{7}\)
Bài 2:
a) S1 = 22 + 42 + ... + 202
=> S1 = 22(1+22+...+102)
=> S1 = 22.385
=> S1 = 1540
b) S2 = 1002 + 2002 + ... + 10002
=> S2 = 1002(1+22+...+102)
=> S2 = 1002.385
=> S2 = 3850000
a) 2 x 2 = 4 3 x 3 = 9
2 x 4 = 8 3 x 5 = 15
2 x 6 = 12 3 x 7 = 21
2 x 8 = 16 3 x 9 = 27
4 x 4 = 16 5 x 5 = 25
4 x 2 = 8 5 x 7 = 35
4 x 6 = 24 5 x 9 = 45
4 x 8 = 32 5 x 3 = 15
b) 200 x 4 = 800 300 x 2 = 600
200 x 2 = 400 300 x 3 = 900
400 x 2 = 800 500 x 1 = 500
100 x 4 = 400 100 x 3 = 300
a = S
u1 = 1
q = 4
n = 200 ?