Thu gọn tổng sau :
a) Q = 1 + 32 + 34 +...+ + 32023
ai nhanh tớ tick
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a) \(A=2+2^2+2^3+...+2^{2017}\)
\(2A=2^2+2^3+2^4+...+2^{2018}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)
\(A=2^{2018}-2\)
b) \(C=1+3^2+3^4+...+3^{2018}\)
\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)
\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)
\(8C=3^{2020}-1\)
\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)
\(Toru\)
B = 1 + 32 + 34 + … + 32018
32.B = 32.( 1 + 32 + 34 + … + 32018)
9B = 32 + 34 + 36 + … + 32020
9B – B = (32 + 34 + 36 + … + 32020) – (1 + 32 + 34 + … + 32018)
8B = 32020 – 1
B = (32020 – 1) : 8.
Vậy B = (32020 – 1) : 8.
a: Tổng các số hạng là:
\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)
Ta có: A+1=2x
\(\Leftrightarrow2x=24311\)
hay \(x=\dfrac{24311}{2}\)
7 . (-2)^3 - 60:15
= 7 . (-8) - 4
= (-56)- 4
= - 60
125 - (-75) + 32 - (48 + 32)
= 125-75 + 32 - 48 -32
= 50 + (-48)
=2
a) \(\left(3^{35}+3^{34}-3^{33}\right):3^{32}\)
\(=\frac{3^{35}}{3^{32}}+\frac{3^{34}}{3^{32}}-\frac{3^{33}}{3^{32}}\)
\(=3^3+3^2-3\)
\(=27+9-3\)
\(=33\)
b) \(5^3.37+5^3.64-5^7:5^4\)
\(=5^3.37+5^3.64-5^3\)
\(=5^3\left(37+64-1\right)\)
\(=5^3.100\)
\(=125.100\)
\(=12500\)
\(\left(3^{35}+3^{34}-3^{33}\right)\div3^{32}=3^{33}\left(3^2+3-1\right)\div3^{32}\)
\(=3^{33}.11\div3^{32}=11\left(3^{33-32}\right)=11.3=33\)
Sửa đề: Q=1+3^2+3^4+...+3^2022
9Q=3^2+3^4+...+3^2024
=>8Q=3^2024-1
=>\(Q=\dfrac{3^{2024}-1}{8}\)