thực hiện tính:
E= 1+\(\frac{1}{2}\) (1+2) + \(\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+\right)+...+\frac{1}{2016}\left(1+2+...+2016\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
12 tháng 11 2016
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+2016\right)\)
\(A=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{16}.\frac{\left(1+16\right).16}{2}\)
\(A=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+\frac{1}{4}.\frac{5.4}{2}+...+\frac{1}{16}.\frac{17.16}{2}\)
\(A=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(A=\frac{1}{2}.\left(2+3+4+5+...+17\right)\)
\(A=\frac{1}{2}.\frac{\left(2+17\right).16}{2}=19.4=76\)
12 tháng 11 2016
hik như vế sau là a làm theo 16 chứ k fai 2016 hay sao ấy
24 tháng 9 2016
\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+..+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)
\(C=1+\frac{1}{2}.\left(1+2\right).2:2+\frac{1}{3}.\left(1+3\right).3:2+\frac{1}{4}.\left(1+4\right).4:2+...+\frac{1}{2016}.\left(1+2016\right).2016:2\)
\(C=1+3:2+4:2+5:2+...+2017:2\)
\(C=2.\frac{1}{2}+3.\frac{1}{2}+4.\frac{1}{2}+5.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(C=\frac{1}{2}.\left(2+3+4+5+...+2017\right)\)
\(C=\frac{1}{2}.\left(2+2017\right).2016:2\)
\(C=\frac{1}{2}.2019.2016.\frac{1}{2}\)
\(C=2019.504=1017576\)
28 tháng 2 2018
\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
28 tháng 2 2018
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
Xét Sn = 1+2+3+4+...+n (1)
=> Sn= n+(n-1)+...+2+1 (2)
Thấy 1+n = 2+(n-1) = 3+(n-2) = n-1+2=n+1
Lấy (1);(2) và chú ý trên ta có:
2.Sn = (n+1)+(n+1)+(n+1)+...+(n+1)=n(n+1) (vì n số hạng giống nhau)
=> Sn= n(n+1)/2 => Sn/n = (n+1)/2
=> P= 1+ S2/2 + S3/3 + S4/4 +...+ Sn/n
P= 1+3/2+4/2+5/2+...+(n+1)/2
P= 2(2+3+4+...+n+n+1) = 2(1+2+...n+n+1) - 2 = 2.S(n+1) - 2
P= 2.(n+1)(n+2)/2 -2 = (n+1)(n+2) -2 = n2+3n
Bài toán chỉ đến S2016/2016 (tức n=2016)
Vậy S= 20162+3.2016=2016.(2016+3)=2016.2019=4070304
E = 1 + 1/2.(1 + 2) + 1/3.(1 + 2 + 3) + 1/4.(1 + 2 + 3 + 4) + ... + 2016.(1 + 2 + 3 + ... + 2016)
E = 1 + 1/2.(1 + 2).2:2 + 1/3.(1 + 3).3:2 + 1/4.(1 + 4).4:2 + ... + 2016.(1 + 2016).2016:2
E = 2/2 + 3/2 + 4/2 + 5/2 + ... + 2017/2
E = 2+3+4+5+...+2017/2
E = (2 + 2017).2016/2
E = 2019.1008
E = 2 035 152