Cho A=\(5-5^2+5^3-5^4+........-5^{98}+5^{99}\) .Tính A
Cho A = 5 - 5^2 + 5^3 - 5^4 + …- 5^98 + 5^99 . Tính tổng A.
Answer:
\(A=5-5^2+5^3-5^4+...-5^{98}+5^{99}\)
`=>5A=5^2-5^3+5^4-5^5+...-5^{99}+5^{100}`
`=>5A+A=(5^2+5^3-5^4+...-5^{98}+5^{99})`
`=>6A=5+5^{100}`
`=>A=\frac{5+5^{100}}{6}`
Cho A = 5 - 5^2 + 5^3 -5^4 +...- 5^98 + 5^99 . Tính A
A = 5/2*3*4 + 5/3*4*5 + ... + 5/98*99*100 + 5/99*100*101
Tính tổng A
1. Cho A=5-5^2+5^3-5^4+...-5^98+5^99
tính tổng A
mik sẽ tick
5A = 52 + 53 + 54 +...+ 5^100
=> 4A = 5A - A = 5^100 - 5 = 5(5^99-1)
=> A = 5(5^99-1)/4
A = 5 – 5^2 + 5^3 – 5^4 + …- 5^98 + 5^99 =>5A = 5^2 – 5^3 + 5^4 - …+ 5^98 – 5^99 + 5^100
Tính và rút gọn được 6A = 5 + 5^100
A=(5+5^100):6
Vậy A=(5+5^100):6
Tính tổng : A =5 - 5^2 + 5^3 - 5^4+.......- 5^98 + 5^99
Bài 1: Tính
a, A = 7 + 7^3 + 7^5 + ...... + 7^151
b, B = 11^4 + 11^5 + 11^6 + ....... + 11^50
c, C = ( 2/3 )^4 + ( 2/3 )^5 + ( 2/3 )^6 + ..... + ( 2/3 )^100 ( 2/3 nghĩa là 2 phần 3 )
d, D = 5^100 - 5^99 - 5^98 - 5^97 -.....- 5^2 - 5 - 1
Bài 2: Cho A = 1 + 4 + 4^2 + ..... + 4^99
B = 4^100
a, Tìm B - A
b, Chứng minh rằng A < B/3 ( B/3 nghĩa là B phần 3 )
Bài 3 : Tính
a, A = 7^2 + 14^2 + 21^2 + 28^2 + ...... + 371^2
b, B = 11^2 + 22^2 + 33^2 + ...... + 1661^2
Bài 4 : Tính
A = 99 x 1 + 98 x 2 + 97 x 3 + ....... + 3 x 97 + 2 x 98 +1 x 99
Cho A=2×9×8+3×12×10+4×15×12+...+98×297×200/23×4+3×4×5+4×5×6+..+98×99×100
Tính a2
A=5-5^2=+5^3-5^4+.....-5^98+5^99
A = 5 - 52 + 53 - 54 +...+ 597 - 598 + 599
5A = 52 - 53 + 54-.....-597 + 598 - 599 + 5100
5A + A = 5 + 5100
6A = 5 + 5100
A = \(\dfrac{5+5^{100}}{6}\)
tính nhanh
a) 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + ......... + 1/98*99 + 1/99*100
b) 5/11*16 + 5/16*21 + 5/21*16 + .......... + 5/61*66
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b)\(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)
\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}\)
\(=\frac{5}{66}\)
a,\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
ta có:
\(\frac{1}{1.2}=\frac{2-1}{1.2}=\frac{2}{1.2}-\frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)
...
\(\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}=\frac{99}{100}\)
b,
\(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.16}+...+\frac{5}{61.66}\)
ta có:
\(\frac{5}{11.16}=\frac{16-11}{11.16}=\frac{16}{11.16}-\frac{11}{11.16}=\frac{1}{11}-\frac{1}{16}\)
\(\frac{5}{16.21}=\frac{21-16}{16.21}=\frac{21}{16.21}-\frac{16}{16.21}=\frac{1}{16}-\frac{1}{21}\)
...
\(\frac{5}{61.66}=\frac{66-61}{61.66}=\frac{66}{61.66}-\frac{61}{61.66}=\frac{1}{61}-\frac{1}{66}\)
= \(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
=\(\frac{1}{11}-\frac{1}{66}\)=\(\frac{5}{66}\)