Rút gọn: a) A= 1/3 + 1/32 + 1/33 +.....+ 1/3100
rút gọn :
A=1+3+32+33+....+3100
B=1+12+24+...+2100
C=1-3+32-33+...+3100
A = 1 + 3 + 32 + 33 + ... + 3100
3A = 3 + 32 + 33 +34+ .... + 3101
3A - A = (3 + 32 + 34 + ... + 3101) - (1 + 3 + 32 + 33 + ... + 3100)
2A = 3 + 32 + 34 + ... + 3101 - 1 - 3 - 32 - 33 - ... - 3100
2A = (3 - 3) + (32 - 32) + ... + (3100 - 3100) + (3101 - 1)
2A = 3101 - 1
A = \(\dfrac{3^{101}-1}{2}\)
Solution
We have: 3A = 3. (1 + 3 + 32 + 33 + ... + 399 + 3100) (1 + 3 + 32 + 33 + ... + 399 + 3100)
3A = 3 + 32 + 33 + ... + 3100 + 31013 + 32 + 33 + ... + 3100 + 3101
Inferred: 3A - A = (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100)
2A = 3101−13101−1
⇒⇒ A = 3101−123101−12
So A = 3101−12
Please help me
Dịch ra là: Ta có: 3A = 3. (1 + 3 + 32 + 33 + ... + 399 + 3100) (1 + 3 + 32 + 33 + ... + 399 + 3100) 3A = 3 + 32 + 33 + ... + 3100 + 31013 + 32 + 33 + ... + 3100 + 3101 Suy ra: 3A - A = (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) ⇒⇒ A = 3101−123101−12 Vậy A = 3101−12
Mà đoạn 2A sai nhé bạn, sửa lại:
2A = 3101−13101−1 2A=-10001
A=-10001/2
A=-5000,5
Vậy A=-5000,5
Bài 1: tính tổng dãy số sau:
A = 1+3+32+33+...+399+3100
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Giải
Ta có: 3A = 3.(1+3+32+33+...+399+3100)(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+31013+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)(3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−13101−1
⇒⇒ A = 3101−123101−12
Vậy A = 3101−12
xin lỗi bài trên của mình làm sai
Ta có: 3A = 3.(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−1
⇒ A = 3101−1
2
Vậy A = 3101−1
2
a) Thu gọn tổng sau A = 1 + 2 + 22 + 23 + ….+ 219 + 220. Tìm x biết A + 1 = 2x
b) Cho B = 1 + 3 + 32 + 33+ …. + 399 + 3100.Tìm x biết 2B + 1 = 3x+1
a: Tổng các số hạng là:
\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)
Ta có: A+1=2x
\(\Leftrightarrow2x=24311\)
hay \(x=\dfrac{24311}{2}\)
Rút gọn 3100-399+398-397+...+32-3+1
\(A=3^{100}-3^{99}+3^{98}-...-3+1\\ \Rightarrow\dfrac{1}{3}A=3^{99}-3^{98}+3^{97}-...-1+\dfrac{1}{3}\\ \Rightarrow\dfrac{4}{3}A=3^{100}+\dfrac{1}{3}\\ \Rightarrow A=\dfrac{3^{101}}{4}+\dfrac{1}{4}\)
thu gọn tổng sau
B=1+3+32+33+...+3100+3101
\(B=1+3+3^2+3^3+...+3^{100}+3^{101}\)
\(\Rightarrow3B=3+3^2+3^3+3^4+...+3^{101}+3^{102}\)
\(\Rightarrow3B-B=3^{102}-1\)
\(\Leftrightarrow2B=3^{102}-1\)
\(\Leftrightarrow B=\dfrac{3^{102}-1}{2}\)
Chứng minh rằng:
A = 1/3 + 1/32 + 1/33 + ..........+ 1/399 < 1/2
B = 3/12x 22 + 5/22 x 32 + 7/32 x 42 +............+ 19/92 x 102 < 1
C = 1/3 + 2/32 + 3/33 + 4/34 +.........+ 100/3100 ≤ 0
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
tính tổng sau : A = 1+3+32+33+...+3100
\(A=1+3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{101}\)
Trừ theo vế:
\(\Rightarrow3A-A=\left(3+3^2+3^3+...3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-1\Rightarrow A=\dfrac{3^{101}-1}{2}\)
tính tổng sau :A =1+3+32 +33 +...+ 3100
A =1+3+32 +33 +...+ 3100
3A=3.(30+3+32 +33 +...+ 3100)
3A=31+32 +33 +...+ 3101
3A-A=(31+32 +33 +...+ 3101)-(30+3+32 +33 +...+ 3100)
2A=3101-30
A=(3101-1) :2
vậy A=(3101-1) :2
t.i.c cho mình nha
Tính A = 1 - 3 + 32 - 33 + 34 - ... + 398 - 399 + 3100
Tham khảo
Ta có: 3A = 3.(1+3+32+33+...+399+3100)(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+31013+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)(3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−13101−1
⇒⇒ A = 3101−123101−12
Vậy A = 3101−12