Tính
A=\(1^2+2^2+3^2+...+100^2\)
B=\(1^3+2^3+3^3+...+100^3\)
K
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23 tháng 10 2016
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
DV
0
DV
1
9 tháng 10 2015
A=1+2+22+…+2100
2A=2(1+2+22+…+2100)
2A=2+22+…+2101
2A-A = A = 2+22+…+2101-(1+2+22+…+2100)
A = 2+22+…+2101-1-2-22-…-2100
A = (2-2)+(22-22)+…+(2100-2100)+2101-1
A = 0+0+…+0+2101-1
A = 2101-1
B=3-32+33-34+…+299-3100
3B = 3(3-32+33-34+…+299-3100)
3B = 32-33+34-…-299+3100-3101
3B+B = 4B = 3-32+33-34+…+299-3100
4B =(3-32+33-34+…+299-3100)+(32-33+34-…-299+3100-3101)
4B =3-32+33-34+…+299-3100+32-33+34-…-299+3100-3101
4B =3+(32-32)+(33-33)+(34-34)+…+(299-299)+(3100-3100)-3101
4B =3+0+0+0+....+0-3101
4B =3-3101
B = (3-3101)/4
DK
1
KT
29 tháng 7 2018
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
\(\Rightarrow\)\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
\(\Rightarrow\)\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow\)\(A=2-\frac{1}{2^{100}}\)
\(B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow\)\(3B=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\)
\(\Rightarrow\)\(3B-B=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow\)\(2B=3-\frac{1}{3^{100}}\)
\(\Rightarrow\)\(B=\frac{3-\frac{1}{3^{100}}}{2}\)
NN
0
TA
1
5 tháng 2 2022
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
a) \(A=1^2+2^2+3^2+...+100^2\)
\(A=1.1+2.2+3.3+...+100.100\)
\(A=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+100.\left(101-1\right)\)
\(A=1.2-1+2.3-2+3.4-3+...+100.101-100\)
\(A=\left(1.2+2.3+3.4+...+100.101\right)-\left(1+2+3+...+100\right)\)
Đặt \(C=1.2+2.3+3.4+...+100.101\)và \(D=1+2+3...+100\)
Ta có:
\(C=1.2+2.3+3.4+...+100.101\)
\(3C=1.2.3+2.3.3+3.4.3+...+100.101.3\)
\(3C=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+100.101\left(102-99\right)\)
\(3C=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\)
\(3C=100.101.102\)
\(3C=1030200\)
\(C=343400\)
Ta lại có:
\(D=1+2+3+...+100\)
\(D=\dfrac{100-1+1}{2}.\left(1+100\right)\)
\(D=50.101\)
\(D=5050\)
\(\Rightarrow A=C-D\)
\(\Rightarrow A=343400-5050=338350\)
b) Ta thấy:
\(\left(n-1\right)n\left(n+1\right)\)
\(=\left(n^2-n\right)\left(n+1\right)\)
\(=\left(n^2-n\right)n+n^2-n\)
\(=n^3-n^2+n^2-n\)
\(=n^3-n\)
\(\Rightarrow n^3=\left(n-1\right)n\left(n+1\right)+n\left(1\right)\)
Áp dụng (1) vào B ta có:
\(B=1^3+2^3+3^3+...+100^3\)
\(B=\left(1-1\right)1\left(1+1\right)+1+\left(2-1\right)2\left(2+1\right)+2+...+\left(100-1\right)100\left(100+1\right)\)
\(B=1+2+1.2.3+3+2.3.4+...+100+99.100.101\)
\(B=\left(1+2+3+...+100\right)+\left(1.2.3+2.3.4+...+99.100.101\right)\)
Đặt \(E=1+2+3+...+100\)và \(F=1.2.3+2.3.4+...+99.100.101\)
Ta có:
\(E=1+2+3+...+100\)
\(E=\dfrac{100-1+1}{2}.\left(1+100\right)\)
\(E=50.101=5050\)
Ta lại có:
\(F=1.2.3+2.3.4+...+99.100.101\)
\(4F=1.2.3.4+2.3.4.4+...+99.100.101.4\)
\(4F=1.2.3.4+2.3.4\left(5-1\right)+...+99.100.101\left(102-98\right)\)
\(4F=1.2.3.4+2.3.4.5-1.2.3.4+...+99.100.101.102-98.99.100.101\)
\(4F=99.100.101.102\)
\(4F=101989800\)
\(F=25497450\)
Vì \(B=E+F\)
\(\Rightarrow B=5050+25497450\)
\(\Rightarrow B=25502500\)