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 \(A=2^{2016}-1\)
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Bài 2 :
\(B=3^{11}+3^{12}+3^{13}+...+3^{101}\)
\(3B=3^{12}+3^{13}+3^{14}+...+3^{102}\)
\(3B-B=\left(3^{12}+3^{13}+3^{14}+...+3^{102}\right)-\left(3^{11}+3^{12}+3^{13}+...+3^{101}\right)\)
\(2B=3^{102}-3^{11}\)
\(B=\frac{3^{102}-3^{11}}{2}\)
Vậy \(B=\frac{3^{102}-3^{11}}{2}\)
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\(B=3^{11}+3^{12}+...+3^{101}\)
\(\Rightarrow3B=3^{12}+3^{13}+3^{14}+...+3^{101}\)
\(\Rightarrow3B-B=3^{12}+3^{13}+3^{14}+...+3^{101}-3^{11}-3^{12}-3^{13}-...-3^{101}\)
\(\Rightarrow2B=3^{101}-3^{11}\)
\(\Rightarrow B=2^{101}-3^{11}:2\)
Bài 3 :
\(C=1+5^2+5^3+...+5^{200}\)
\(5C=5+5^3+5^4+...+5^{201}\)
\(5C-C=\left(5+5^3+5^4+...+5^{201}\right)-\left(1+5^2+5^3+...+5^{200}\right)\)
\(4C=5^{201}+5-5^2-1\)
\(4C=5^{201}-21\)
\(C=\frac{5^{201}-21}{4}\)
Vậy \(C=\frac{5^{201}-21}{4}\)
Chúc bạn học tốt ~
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