C=3+3^2+3^5+...+3^99
a, A= 3 + 32 + 33 + ..... + 399
b, B=3+33+ 35+....+399
c, C=1+2 +22 + ....... + 299
d, D= 1+ 22 +24 + ...... + 298
e, E= 5+ 53 + 55 +.... + 599
A=3 + 3^2 + 3^3 +...+3^99
3.A= 3^2 + 3^3 +...+3^99 + 3^100
2.A= 3^100 - 3
A= (3^100 - 3) : 2
làm vậy có đúng ko bn?
bài 1
A=1*2*3+2*3*4+3*4*5+...+99*100*101
B=1*3*5+3*5*7+...+95*97*99
C=2*4+4*6+..+98*100
D=1*2+3*4+5*6+...+99*100
E=1^2+2^2+3^2+...+100^2
G=1*3+2*4+3*5+4*6+...+99*101+100*102
H=1*2^2+2*3^2+3*4^2+...+99*100^2
I=1*2*3+3*4*5+5*6*7+7*8*9+...+98*99*100
K=1^2+3^2+5^2+...+99^2
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
Tính
A=1*2+2*3+3*4+..+99*100
B=1*3+3*5+5*7+..+97*99
C=1*2*3+2*3*4+...+98*99*100
A= 1.2 + 2.3 + 3.4 + ... + 99.100
3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3A = 98.99.100
A = 970200 : 3
A = 32340.
a,x/3=2/3+-1/7;b, 2/3*x-4/3=-3/10 c, (2/1*3+2/3*5+2/5*7+ ....+2/97*99) -x=-100/99
tính
a=1*2+2*3+3*4+....+99*100
b=1*3+3*5+5*7+....97*99
c=1*2*3+2*3*4+...98*99*100
giải dùm mk nhoa
A=1+3+3^2+3^3+...+3^100
B=1 phần 2+(1 phần 2)^2+(1 phần 2)^3+...+(1 phần 2)^99
C=5^100-5^99+5^98-5^97+...+5^2-5+1
a
\(A=1+3+3^2+3^3+....+3^{100}\)
\(3A=3+3^2+3^3+3^4+.....+3^{101}\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(B=1-\frac{1}{2^{99}}\)
c
\(C=5^{100}-5^{99}+5^{98}-5^{97}+....+5^2-5+1\)
\(5C=5^{101}-5^{100}+5^{99}-5^{98}+....+5^3-5^2+5\)
\(6C=5^{101}+1\)
\(C=\frac{5^{101}+1}{6}\)
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow\frac{1}{2}B=\)\(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow B-\frac{1}{2}B=\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\right]-\left[\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\Rightarrow\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\Rightarrow B=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)
Tính tổng
A=\(1^3+2^3+3^3+...+100^3\)
B=\(2^3+4^3+...+98^3\)
C=\(1^3+3^3+5^3+...+99^3\)
D=\(1^3-2^3+3^3-4^3+...+99^3-100^3\)
a) Ta có: \(A=1^3+2^3+3^3+...+100^3\)
\(=\left(1-1\right)\cdot1\cdot\left(1+1\right)+1+\left(2-1\right)\cdot2\cdot\left(2+1\right)+2+...+\left(100-1\right)\cdot100\cdot\left(100+1\right)+100\)
\(=1+2+1\cdot2\cdot3+...+99\cdot100\cdot101\)
\(=5050+25497450\)
\(=25502500\)
Chứng minh :
A = 5 + 5^2 + 5^3 + . . . + 5^99 + 5^100 chia hết cho 6
B = 2 + 2^2 + 2^3 + . . . + 2^99 + 2^100 chia hết cho 31
C = 3 + 3^2 + 3^3 + . . . + 3^60 chia hết cho 4, cho 13
A=5+52+...+599+5100
=(5+52)+...+(599+5100)
=5.(1+5)+...+599.(1+5)
=5.6+...+599.6
=6.(5+...+599) chia hết cho 6 (dpcm)
Ccá câu khcs bạn cứ dựa vào câu a mà làm vì cách làm tương tự chỉ hơi khác 1 chút thôi
Chúc bạn học giỏi nha!!
\(A=5+5^2+5^3+...+5^{100}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...\left(5^{99}+5^{100}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{99}\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{99}.6\)
\(=6\left(5+5^3+...+5^{99}\right)⋮6\)(đpcm)
\(B=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+...+2^{96}.31\)
\(=31\left(2+...+9^{96}\right)⋮31\)(đpcm)
\(C=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
\(=3.4+3^3.4+...+3^{59}.4\)
\(=4\left(3+3^3+...+3^{59}\right)⋮4\)(đpcm)
\(C=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=3.13+...+3^{58}.13\)
\(=13\left(3+...+3^{58}\right)⋮13\)(đpcm)
Tính :
A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 99 . 100
B = 1 . 3 + 2 . 4 + 3 . 5 + ... + 99 . 100
C = 1 . 4 + 2 . 5 + 3 . 6 + ... + 99 . 102
D = 4 + 12 + 24 + ... + 19404 + 19800
E = 1 + 3 + 6 + 10 + ... + 4851 + 4950
Các bạn nhớ ủng hộ cho nhất sông núi nhé
Cảm ơn bạn
A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 99 .100
3 . A = 1. 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 99 . 100 . 3
3 . A = 1 . 2 . 3 + 2 . 3 . ( 4 - 1 ) + 3 . 4 . ( 5 - 2 ) + ... + 99 . 100 . ( 1001 - 998 )
3 . A = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 99 . 100 . 1001 - 998 . 99 . 100
3 . A = 99 . ( 100 . 10 )
A = ( 99 . 100 . 10 ) : 3
A = 33000
Tính B=1*3+5*7+9*11+...+97*101
C=1*3*5-3*5*7+5*7*9-....-97*99*101
D=1*99+3*97+5*95+...+49*51
E=1*3^3+3*5^3+5*7^3+...+49*51^3
F=1*99^2+2*98^2+3*97^2+...+49*51^2
cái này bạn mở sách bồi dưỡng toán ra trang gần cuối là thấy ngay ấy mà