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\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{49}+\left(-5\right)^{50}\\ -5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+\left(-5\right)^4+...+\left(-5\right)^{50}+\left(-5\right)^{51}\\ B+5B=\left[\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{49}+\left(-5\right)^{50}\right]-\left[\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+\left(-5\right)^4+...+\left(-5\right)^{50}+\left(-5\right)^{51}\right]\\ 6B=\left(-5\right)^0-\left(-5\right)^{51}\\ B=\frac{1-\left(-5\right)^{51}}{6}\)
A = 1 + 5 + 52 + 53 + 53 + ...+ 549 + 550
5A = 5(50+51+52+53+...+549+550)
5A=51+52+53+54+...+550+551
5A-A=(51+52+53+54+...+550+551)-(50 + 51 + 52 + 53 + 53 + ...+ 549 + 550)
4A=551-1
A=(551-1):4
5A = 5 + 5^2 + 5^3 + 5^4 + 5^5 +...+ 5^50 + 5^51
=> 4A = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 +...+ 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 +...+ 5^49 + 5^50 )
=> 4A = 5^51 - 1
=> A = \(\frac{5^{51}-1}{4}\)
5A=5+52+53+...+550+551
5A-A=551-1
A=551-1:4
tick mk nha cái kia sai rôi
Ta có :A = 1 + 5 + 52 + 53 + .... + 549 + 550
=> 5A = 5 + 52 + 53 + .... + 550 + 551
=> 5A - A = 551 - 1
=> 4A = 551 - 1
=> A = \(\frac{5^{51}-1}{4}\)
\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{43}}\)
\(A=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}\)
\(A=\frac{5^{30}.7^{48}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}=5.7^3.\left(1-7.2^2\right)=1715.\left(-27\right)=-46305\)
\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}\left(2^4\right)^2.7^{43}}=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}=\frac{7^{48}.5^{30}.2^8\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}\)
=\(7^5.5.\left(-27\right)=-2268945\)
Ta có:
A = 1+ 5 + 52 + 53 + ......... + 549 + 550
=> 5A = 5 + 52 + 53 + 54 +.......+ 549 + 550
Do đó: 5A - A = 551 - 1
Vậy A = \(\frac{5^{51}-1}{4}\)
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
\(5A=5^1+5^2+5^3+5^4+...+5^{51}\)
\(4A=5A-A=5^{51}-1\)
\(\Rightarrow A=\frac{5^{51}-1}{4}\)
b/
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(\frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{100}\)
\(\frac{1}{2}B=B-\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\)
\(B=\frac{1}{2}B\cdot2=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)
\(B=1-\frac{1}{2^{99}}\)
Rút gọn:
\(A=5^0+5^1+5^2+...+5^{99}+5^{50}\)
\(5A=5^1+5^2+5^3+...+5^{51}\)
\(5A-A=\left(5^1+5^2+5^3+...+5^{51}\right)-\left(5^0+5^1+5^2+...+5^{50}\right)\)
\(4A=5^{51}-5^0\)
\(=>A=\left(5^{51}-5^0\right):4\)
Vậy : \(A=\left(5^{51}-5^0\right):4\)