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5 thành 51
1+1+2+3+4+.......+49+50 rồi tính số số hạng,tìm tổng.cuối cùng +1
A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90
2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100
2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - ( 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 )
A = 2^100 - 2^3
B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50
5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51
5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )
4B = 5^51 - 1
B = 5^51 - 1 / 4
\(a,\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
\(=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{132}.\dfrac{132}{103}=\dfrac{115}{103}\)
\(b,\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2-5^0\)
\(=\dfrac{1}{4}-4-1=-\dfrac{19}{4}\)
\(c,12\dfrac{1}{3}-\dfrac{5}{7}:\left(24-23\dfrac{5}{7}\right)\)
\(=\dfrac{37}{3}-\dfrac{5}{7}.\dfrac{2}{7}=\dfrac{37}{3}-\dfrac{10}{49}=\dfrac{1783}{147}\)
Đặt
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
\(=>5A=5+5^2+5^3+5^4+...+5^{50}+5^{51}\)
\(=>5A-A=\left(5+5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{49}+5^{50}\right)\)
\(=>5A-A=5^{51}-1=>4A=5^{51}-1=>A=\frac{5^{51}-1}{4}\)
Vậy............
Đặt A = 1 + 5 + 52 + 53 + 54 + ..... + 549 + 550
5A = 5 + 52 + 53 + 54 + 55 + ........ + 550 + 551
=> 5A - A = 551 - 5
4A = 551 - 5
A = (551 - 5) : 4
Ủng hộ mk nha!!!
B=1+5+5^2+5^3 + ...5^50
=> 5B=5+5^2+5^3+...+5^51=> 5B-B=4B=(5+5^2+5^3+...+5^51)-(1+5+5^2+5^3 + ...5^50)=> 4B=5^51-1=> B=\(\frac{5^{51}-1}{4}\)\(B=1+5+5^2+5^3+...+5^{50}\)
\(\Rightarrow5B=5+5^2+5^3+...+5^{51}\)
\(\Rightarrow5B-B=\left(5+5^2+5^3+...+5^{51}\right)-\left(1+5+5^2+...+5^{50}\right)\)
\(\Rightarrow4B=5^{51}-1\)
\(\Rightarrow B=\frac{5^{51}-1}{4}\)
\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)
\(A=2^{21}-2\)
B tương tự câu A
\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)
\(C=\dfrac{5^{51}-5}{4}\)
\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)
\(D=\dfrac{3^{101}-1}{2}\)
\(A=2^1+2^2+2^3+...+2^{20}\)
\(2\cdot A=2^2+2^3+2^4+...+2^{21}\)
\(A=2^{21}-2\)
\(B=2^1+2^3+2^5+...+2^{99}\)
\(4\cdot B=2^3+2^5+2^7+...+2^{101}\)
\(B=\)\(\left(2^{101}-2\right):3\)
\(C=5^1+5^2+5^3+...+5^{50}\)
\(5\cdot C=5^2+5^3+5^4+...+5^{51}\)
\(C=(5^{51}-5):4\)
\(D=3^0+3^1+3^2+...+3^{100}\)
\(3\cdot D=3^1+3^2+3^3+...+3^{101}\)
\(D=(3^{101}-1):2\)