Tính (1/2)^2+(1/2)^3+(1/2)^4+...+(1/2)^98+(1/2)^99+(1/2)^99
Những câu hỏi liên quan
tính nhanh (2/3+3/4+5/6+...+99/100).(1/2+2/3+3/4+...+98/99)-(1/2+1/3+...+99/100).(2/3+2/4+...+98/99)
Tính nhanh
a.(1+1/2)×(1+1/3)×(1+1/4)×...×(1+1/98)×(1+1/99).
b.1/2×2/3×3/4×...×97/98×98/99×99/100
a,=3/2*4/3*....100/99
=3*4*5*....*100/2*3*...*99
=100/2=50
b, nhân lên băng:
1*2*3*...*99/2*3*...*100=1/100
1. Tính
A=1*3+2*4+.....................................+97*99+98*100
B=1*2*3+2*3*4+...............................+48*49*50
C=12 +22 + 32+................................+982
D=1*99+2*98+......................................+99*1
Ta thấy:
\(A=1\cdot3+2\cdot4+...+97\cdot99+98\cdot100\)
\(A=1\cdot\left(1+2\right)+2\cdot\left(1+3\right)+...+97\cdot\left(1+98\right)+98\cdot\left(1+99\right)\)
\(A=\left(1+1\cdot2\right)+\left(2+2\cdot3\right)+...+\left(97+97\cdot98\right)+\left(98+98\cdot99\right)\)
\(A=\left(1+2+...+97+98\right)+\left(1\cdot2+2\cdot3+...+97\cdot98+98\cdot99\right)\)
Đặt \(B=1+2+...+97+98\) ; \(C=1\cdot2+2\cdot3+...+97\cdot98+98\cdot99\). Khi đó: \(A=B+C\)
* Do số các số hạng của tổng B là: ( 98 - 1 ) : 1 + 1 = 98 ( số hạng ) nên:
\(B=1+2+...+97+98=\frac{\left(98+1\right)\cdot98}{2}=99\cdot49=4851\)
* Ta thấy:
\(C=1\cdot2+2\cdot3+...+97\cdot98+98\cdot99\)
\(\Rightarrow3\cdot C=1\cdot2\cdot3+2\cdot3\cdot3+...+97\cdot98\cdot3+98\cdot99\cdot3\)
\(\Rightarrow3\cdot C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+97\cdot98\cdot\left(99-96\right)+98\cdot99\cdot\left(100-97\right)\)
\(\Rightarrow3\cdot C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+97\cdot98\cdot99-96\cdot97\cdot98+98\cdot99\cdot100-97\cdot98\cdot99\)
\(\Rightarrow3\cdot C=98\cdot99\cdot100\)
\(\Rightarrow C=\frac{98\cdot99\cdot100}{3}\)
\(\Rightarrow C=98\cdot33\cdot100\)
\(\Rightarrow C=323400\)
Vậy: \(A=B+C=4851+323400=328251\)
Đúng 0
Bình luận (0)
Cho A = 100+1/99+2/98+...99/1
B = 100-1/2-2/3-3/4-...-98/99
Tính A/B
Tính A:B
a)A=98+1/2+1/3+1/4+...+1/99
B=2/3+4/3+5/4+...+100/99
b)A=2018+1/2+1/3+1/4+...+1/2019
B=3/2+4/3+3/4+...+2020/2019
c)A=99/1+98/2+97/3+...+2/98+1/99
B=1/2+1/3+1/4+...+1/100
Giải đầy đủ
a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)
b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{2019}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)
c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)
\(A=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)
Đúng 0
Bình luận (0)
a)\(B=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{100}{99}\)
\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{99}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\right)\)
\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)
\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)
Vậy \(A:B=1.\)
c)\(A=\left(1+1+...+1\right)+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)
\(A=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+...+\left(1+\frac{2}{98}\right)+\left(1+\frac{1}{99}\right)\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
Đúng 0
Bình luận (0)
Tính B = 1/2+(1/2)^2+ (1/2)^3+(1/2)^4+....+(1/2)^98+(1/2)^99+(1/2)^99 Ta được B=
Ta có: B= \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
=> \(\frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\)
=> B - \(\frac{1}{2}B=\left(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\right)\)
\(-\left(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{4}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\right)\)
=> B - \(\frac{1}{2}B=\left(\frac{1}{2}+\left(\frac{1}{2}\right)^{99}\right)-\left(\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\right)=\frac{1}{2}\)
=> B \(\times\left(1-\frac{1}{2}\right)=\frac{1}{2}\)
=> B = 1
Câu này chắc chắn đúng luôn
Đúng 0
Bình luận (0)
Bài 4: Tính tổng 1) 1 + (-2) + 3 + (-4) + . . . + 19 + (-20) 2) 1 – 2 + 3 – 4 + . . . + 99 – 100 3) 2 – 4 + 6 – 8 + . . . + 48 – 50 4) – 1 + 3 – 5 + 7 - . . . . + 97 – 99 5) 1 + 2 – 3 – 4 + ... + 97 + 98 – 99 - 100
1. 1 + ( -2) +3 +(-4) + .........+ 19 + (-20)
= -1 + ( -1) +....+(-1)
= -1. 10
= -10
2. 1 – 2 + 3 – 4 + . . . + 99 – 100
= ( -1) + (-1) +....+(-1)
= -1. 50
= -50
3. 2 – 4 + 6 – 8 + . . . + 48 – 50
= (-2) + (-2) +....+ (-2)
= -2. 12 + 26
= -24 + 26
= 2
4. – 1 + 3 – 5 + 7 - . . . . + 97 – 99
= 2 + 2 +......+2
= 2.25
= 50
5. 1 + 2 – 3 – 4 + ... + 97 + 98 – 99 - 100
= (1+2-3-4) +......+ ( 97+98-99 -100)
= -4 . (-4).....(-4)
= -4. 25
= -100
Tính S = [ 1 + (1+2) + (1+2+3) +...+ (1+2+3+...+98) ] / (1*2 + 2*3 + 3*4 +...+ 98*99)
\(S=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+....+98.99}\)
\(=\frac{1+\frac{2.3}{2}+\frac{3.4}{2}+....+\frac{98.99}{2}}{1.2+2.3+3.4+....+98.99}\)
\(=\frac{\frac{1}{2}\left(1.2+2.3+3.4+....+98.99\right)}{1.2+2.3+3.4+....+98.99}\)
\(=\frac{1}{2}\)
Đúng 0
Bình luận (0)
sao toàn thấy frac, left và right ko vậy?
Đúng 0
Bình luận (0)
tính
1 : 99 / 100 : 98 / 99 : 97 / 98 : ... : 3 / 4 : 2 / 3 : 1 / 2
\(=\frac{99}{100}.\frac{99}{98}.\frac{98}{97}.\frac{97}{96}.....\frac{4}{3}.\frac{3}{2}.\frac{2}{1}\)
Ta loại các số giống nhau ở tử và mẫu thì được
\(\frac{99}{100}.\frac{99}{1}\)
\(=\frac{9801}{100}\)
Đúng 0
Bình luận (0)
= \(\frac{99}{100}.\frac{99}{98}.\frac{98}{97}.\frac{96}{97}...\frac{4}{3}.\frac{3}{2}.\frac{2}{1}\)
Ta loại các số giống nhau ở tử số và mẫu số thì đc :
\(\frac{99}{100}.\frac{99}{1}\)
= \(\frac{9801}{100}\)
Đúng 0
Bình luận (0)