Cho \(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2016}{4^{2016}}\)
So Sánh S với \(\frac{1}{2}\)
so sánh: S=\(\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{2016}{2017!}\)
Ta có :
\(S=\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{2016}{2017!}\)
\(S=\frac{3-1}{3!}+\frac{4-1}{4!}+\frac{5-1}{5!}+...+\frac{2017-1}{2017!}\)
\(S=\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+\frac{5}{5!}-\frac{1}{5!}+...+\frac{2017}{2017!}-\frac{1}{2017!}\)
\(S=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{4!}-\frac{1}{5!}+...+\frac{1}{2016!}-\frac{1}{2017!}\)
\(S=\frac{1}{2!}-\frac{1}{2017!}\)
\(S=\frac{1}{2}-\frac{1}{2017!}\)
Vậy \(S=\frac{1}{2}-\frac{1}{2017!}\)
Chúc bạn học tốt ~
Àk mình quên còn so sánh với \(\frac{1}{2}\) nữa bạn thêm vào nhé
\(S=\frac{1}{2}-\frac{1}{2017!}< \frac{1}{2}\)
Vậy \(S< \frac{1}{2}\)
Chúc bạn học tốt ~
Người ta tròng 12 cây thành 7 hàng , mỗi hàng có 4 cây hãy vẽ sơ đồ vị trí các cây .
Cho s = \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+.........+\frac{2015}{2^{2016}}\)
So sánh S với 1
ĐANG CẦN GẤP
Cho \(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2016}{4^{2016}}\)
Chứng minh rằng : \(S<\frac{1}{2}\)
Chứng minh : \(S=\frac{1}{4^1}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2016}{4^{2016}}>\frac{1}{2}\)
SO SÁNH:
A=\(\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2016}+\frac{1}{2017}}\)
VÀ
B=2017
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
1/Tính nhanh
P=(1-\(\frac{1}{2^2}\)) x (1-\(\frac{1}{3^2}\)) x (1-\(\frac{1}{4^2}\)) x ... x (1-\(\frac{1}{50^2}\))
2/Cho Q=(1-\(\frac{1}{2^2}\)) x (1-\(\frac{1}{3^2}\)) x (1-\(\frac{1}{4^2}\)) x ... x (1-\(\frac{1}{40^2}\)) . So sánh Q với \(\frac{1}{2}\)
3/So sánh: A = \(\frac{2016^{2016}+1}{2016^{2017}+1}\)và B = \(\frac{2016^{2017}-3}{2016^{2018}-3}\)
P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
So sánh \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2016}{3^{2016}}\)với \(\frac{3}{4}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}=1-\frac{1}{2016}=\frac{2015}{2016}< \frac{1512}{2016}=\frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Cho tổng gồm 2016 số hạng : S= \(\frac{1}{4}\)+ \(\frac{2}{4^2}\)+ \(\frac{3}{4^3}\)+ \(\frac{4}{4^4}\)+ ....... + \(\frac{20166}{4^{2016}}\). Chứng minh rằng: S < \(\frac{1}{2}\)
Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\) Chứng tỏ S < 1
Ta có:
S = 1/22 + 1/32 + 1/42 + ... + 1/20162
= 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/2016.2016
S < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2015.2016
S < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2015 - 1/2016
S < 1 - 1/2016
Mà 1 - 1/2016 < 1
=> S < 1
Vậy S < 1
Ủng hộ nha