Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(A=1+2+2^2+2^3+....+2^{15}\)
\(\Rightarrow2A=2+2^2+2^3+.....+2^{16}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{16}\right)-\left(1+2+2^2+...+2^{15}\right)\)
\(\Rightarrow A=2^{16}-1\)
Câu b đêm nhân lên 3B ( tương tự)
c đêm nhân lên 4C (tương tự)
Câu
Bài 4 :
\(D=11+11^2+11^3+...+11^{1000}\)
\(11D=11^2+11^3+11^4+...+11^{1001}\)
\(11D-D=\left(11^2+11^3+11^4+...+11^{1001}\right)-\left(11+11^2+11^3+...+11^{1000}\right)\)
\(10D=11^{1001}-11\)
\(D=\frac{11^{1001}-11}{10}\)
Vậy \(D=\frac{11^{1001}-11}{10}\)
Chúc bạn học tốt ~
Bài 1 :
\(A=1+2+2^2+....+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)\)
\(A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
Chúc bạn học tốt ~
a ) \(A=2^0+2^1+2^2+...+2^{2010}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2011}\)
\(\Rightarrow2A-A=\left(2+...+2^{2011}\right)-\left(2^0+2^1+...+2^{2010}\right)\)
\(\Rightarrow2A-A=2^{2011}-2^0\)
\(\Rightarrow A=2^{2011}-1\)
b ) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(\Rightarrow3B-B=\left(3+3^2...+3^{2011}\right)-\left(1+3+...+3^{2010}\right)\)
\(\Rightarrow2B=3^{2011}-1\)
\(\Rightarrow B=\frac{3^{2011}-1}{2}\)
Chúc bạn học tốt !!!
a) \(A=2+2^2+2^3+2^4+.....+2^{98}+2^{99}\)
\(\Rightarrow2A=2^2+2^3+2^4+2^5.....+2^{99}+2^{100}\)
\(\Rightarrow2A-A=\left(2^2+2^3+2^4+2^5.....+2^{99}+2^{100}\right)-\left(2+2^2+2^3+2^4+.....+2^{98}+2^{99}\right)\)
\(\Rightarrow A=2^{100}-2\)
b) \(B=2+2^4+2^7+......+2^{97}+2^{100}\)
\(\Rightarrow2^3B=2^4+2^7+......+2^{100}+2^{103}\)
\(\Rightarrow8.B-B=\left(2^4+2^7+......+2^{100}+2^{103}\right)-\left(2+2^4+2^7+......+2^{97}+2^{100}\right)\)
\(\Rightarrow7B=2^{103}-2\)
\(\Rightarrow B=\dfrac{2^{103}-2}{7}\)
Bài 1:
\(2B=2^2+2^3+2^4+2^5+...+2^{101}\\ \Rightarrow2B-B=2^{101}-2\\ \Leftrightarrow B=2^{101}-2\)
\(3C=3+3^2+3^3+3^4+...+3^{2004}\\ \Rightarrow3C-C=3^{2004}-3\\ \Leftrightarrow2C=3^{2004}-3\\ \Leftrightarrow C=\frac{3^{2004}-3}{2}\)
Mấy câu sau tương tự nhân 4 và 5 nhé bạn!
Bài 2: Giải theo lớp 6 nhé! :) Mình nghĩ đề bài cần a nguyên nữa nhé nếu không giải theo lớp 8,9 mất rồi! :)
\(a,2a+27⋮2a+1\\ \Leftrightarrow2a+1+26⋮2a+1\\ \Rightarrow26⋮2a+1\left(vì2a+1⋮2a+1\right)\\ \Rightarrow2a+1\inƯ_{\left(26\right)}mà2a+1lẻnên:\\ 2a+1\in\left\{1;-1;13;-13\right\}\\ \Leftrightarrow a\in\left\{0;-1;6;-7\right\}\\ Vậy...\)
Mấy bài sau tương tự nhé! :)
1,\(A=\)\(1+2+2^2+2^3+...+2^{2015}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2016}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}\right)-\left(1+2+2^2+2^3+...+2^{2015}\right)\)
\(A=\)\(2^{2016}-1\)
~~~Hok tốt~~~
2,\(B=3^{11}+3^{12}+3^{13}+...+3^{101}\)
\(\Rightarrow3B=3^{12}+3^{13}+3^{14}+...+3^{102}\)
\(\Rightarrow3B-B=\left(3^{12}+3^{13}+3^{14}+...+3^{102}\right)-\left(3^{11}+3^{12}+3^{13}+...+3^{101}\right)\)
\(\Rightarrow2B=3^{102}-3^{11}\)
\(\Rightarrow B=\frac{3^{102}-3^{11}}{2}\)
~~~Hok tốt~~~
b, 2B=32+33+34+35+36
2B-B=3+36