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Mk làm mẫu câu a nha
4B = 2^2+2^4+.....+2^202
3B = 4B - B = (2^2+2^4+.....+2^202) - (1+2^2+2^4+.....+2^200) = 2^202 - 1
=> B = (2^202-1)/3
Tk mk nha
Tham khảo ở đây : https://olm.vn/hoi-dap/question/636277.html
a: \(4B=2^2+2^4+...+2^{202}\)
\(\Leftrightarrow3B=2^{202}-1\)
hay \(B=\dfrac{2^{202}-1}{3}\)
b: \(25C=5^3+5^5+...+5^{103}\)
\(\Leftrightarrow24C=5^{103}-5\)
hay \(C=\dfrac{5^{103}-5}{24}\)
cơ bản giống nhau nhé
Lời giải cho ý (D) các câu khác tương tự
\(D=13^1+13^3+13^5+...+13^{99}\)
\(13^2.D=13^3+13^5+13^7...+13^{99+2}\)
\(\left(13^2-1\right)D=13^3+13^5+13^7+...+13^{101}-\left(13^1+13^3+13^5+...+13^{99}\right)\)
\(D=\dfrac{13^{101}-13}{13^2-1}\)
\(A=2+2^3+2^5+2^7+2^9+...+2^{2009}\)
\(\Leftrightarrow\)\(4A=2^3+2^5+2^7+2^9+2^{11}+...+2^{2011}\)
\(\Leftrightarrow\)\(4A-A=\left(2^3+2^5+2^7+...+2^{2011}\right)-\left(2+2^3+2^5+...+2^{2009}\right)\)
\(\Leftrightarrow\)\(3A=2^{2011}-2\)
\(\Leftrightarrow\)\(A=\frac{2^{2011}-2}{3}\)
Ta có :
\(A=2+2^3+2^5+...+2^{2009}\)
\(4A=2^3+2^5+2^7+...+2^{2011}\)
\(4A-A=\left(2^3+2^5+2^7+...+2^{2011}\right)-\left(2+2^3+2^5+...+2^{2009}\right)\)
\(3A=2^{2011}-2\)
\(A=\frac{2^{2011}-2}{3}\)
Vậy \(A=\frac{2^{2011}-2}{3}\)
Câu b) dễ hơn nữa làm tương tư câu a) nhưng B nhân cho 2
Câu c) thì C nhân cho 5
Câu d) thì D nhân cho 169
Mời bạn tham khảo các link sau:
a),b),c):https://hoidap247.com/cau-hoi/214111
d):https://olm.vn/hoi-dap/detail/78449788871.html
\(A=1+7+7^2+7^3+...+7^{200}\)
\(\Rightarrow7A=7+7^2+7^3+...+7^{201}\)
\(\Rightarrow7A-A=\left(7+7^2+...+7^{201}\right)-\left(1+7+7^2+...+7^{200}\right)\)
\(\Rightarrow6A=7^{201}-1\)
\(\Rightarrow A=\frac{7^{201}-1}{6}\)
\(B=5^1+5^3+5^5+...+5^{101}\)
\(\Rightarrow5^2B=5^3+5^5+5^7+...+5^{103}\)
\(\Rightarrow25B-B=\left(5^3+5^5+...+5^{103}\right)-\left(5+5^3+...+5^{101}\right)\)
\(\Rightarrow24B=5^{103}-5\)
\(\Rightarrow B=\frac{5^{103}-5}{24}\)
\(D=1+a+a^2+a^3+...+a^n\)
\(\Rightarrow aD=a+a^2+a^3+...+a^{n+1}\)
\(\Rightarrow aD-D=\left(a+a^2+...+a^{n+1}\right)-\left(1+a+a^2+...+a^n\right)\)
\(\Rightarrow\left(a-1\right)D=a^{n+1}-1\)
\(\Rightarrow D=\frac{a^{n+1}-1}{a-1}\)
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~~~
Mk lầm mẫu câu a nha
Có : 4B = 2^2+2^4+.....+2^202
3B = 4B - B = (2^2+2^4+.....+2^202) - (1+2^2+2^4+....+2^200)
= 2^202 - 1
=> B = (2^202-1)/3
Tk mk nha