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B = 1 + 4 + 42 +...+ 4200 + 4201
=> 4B = 4 + 42 +43 +...+ 4201 + 4202
=> 4B-B = 4202 - 1
3B = 4202 -1
\(\Rightarrow B=\frac{4^{202}-1}{3}\)
4B = 4 + 4^2 + 4^3 + ... + 4^202
4B - B = ( 4 + 4^2 + 4^3 + ... + 4^202 ) - ( 1 + 4 + 4^2 + ... + 4^201 )
3B = 4^202 - 1
B = \(\frac{4^{202}-1}{3}\)
Bạn có chép sai đề bài k ?? sao lại 4 + 4 mũ 3 mà ở cuối lại mà 4 mũ 200 + 4 mũ 201
a) \(D=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7D=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)
\(\Rightarrow7D-D=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6D=1-\frac{1}{7^{100}}\)
\(\Rightarrow D=\left(1-\frac{1}{7^{100}}\right).\frac{1}{6}\)
S=30+32+34+36+...+32020
32.S=32+34+36+...+32020+32021
9S-S=(32+34+36+...+32020+32021)-(30+32+34+36+...+32020)
8S=32021-30
\(S=\frac{3^{2021}-1}{8}\)
\(S=1+2+2^2+...+2^{99}\)
\(S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)
\(S=3+2^2.3+...+2^{98}.3\)
\(=3\left(1+2^2+...+2^{98}\right)⋮3\)
2S=2.(22 + 23 + 24+ ... + 22017 + 22018)
2S=23 + 24+ ... + 22017 + 22018+22019
S=23 + 24+ ... + 22017 + 22018+22019-22 + 23 + 24+ ... + 22017 + 22018
S=22019-22
1/ tính :
a/ A = 341 . 67 + 341 . 16 + 659 . 83
A = 341 . ( 67 + 16 ) + 659 . 83
A = 341 . 83 + 659 . 83
A = 83 . ( 341 + 659 )
A = 83 . 1000
A = 83 000
b/ B = 42 . 53 + 47 . 156 - 47 . 114
B = 42 . 53 + 47 . ( 156 - 114 )
B = 42 . 53 + 47 . 42
B = 42 . ( 53 + 47 )
B = 42 . 100
B = 4 200
2/ thu gọn tổng :
A = 3 + 32 + 33 + ... + 3100
3A = 3^2 + 3^3 + 3^4 + ...+ 3^101
3A - A = ( 3^2 + 3^3 + 3^4 + ...+ 3^101 ) - ( 3 + 32 + 33 + ... + 3100 )
2A = 3^101 - 3
A = 3^101 - 3 / 2
Bài 1:
\(A=341.67+341.16+659.83.\)
\(=341.\left(67+16\right)+659.83\)
\(=341.83+659.83\)
\(=83.\left(341+659\right)\)
\(=83.1000=83000\)
\(B=42.53+47.156-47.114\)
\(=42.53+47.\left(156-114\right)\)
\(=42.53+47.42\)
\(=42.\left(47+53\right)\)
\(=42.100=4200\)
Bài 2:
\(A=3+3^2+3^3+3^4+....+3^{100}\)
\(\Rightarrow3A=3^2+3^3+3^4+3^5+...+3^{101}\)
\(2A=3A-A=\left(3^2+3^3+3^4+3^5+....+3^{101}\right)-\left(3+3^2+3^3+3^4+....+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow A=\frac{3^{101}-3}{2}\)
Bài 3:
\(S=1+2+3+4+...+2018\)
\(=\frac{\left[1+2018\right].\left[\left(2018-1\right)+1\right]}{2}\)
\(=\frac{2019.2018}{2}=2037171\)
\(P=1+3+5+7+...+2017\)
\(=\frac{\left[2017+1\right].\left[\left(2017-1\right):2+1\right]}{2}\)
\(=\frac{2018.1009}{2}\)
\(=1018081\)
Bài 4:
\(\text{Vì ab = 0 }\)\(\Rightarrow\)\(a=0\)\(\text{hoặc}\)\(b=0\)
\(\text{Th1 : ( a = 0)}\)
\(a+4b=16\)
\(0+4b=16\)
\(4b=16\Leftrightarrow b=4\)
\(\text{Th2: ( b = 0)}\)
\(a+4b=16\)
\(a+4.0=16\)
\(a+0=16\Leftrightarrow a=16\)
\(\text{Vậy :}\)\(a;b\in\left\{0;4\right\};\left\{16;0\right\}\)
Bài 5:
\(A=\frac{10^2+11^2+12^2}{13^2+14^2}=\frac{365}{365}=1\)
\(B=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(12.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{12^2.2^{32}}{11.2^{13}.4^{11}-16^9}=....=2\)
a \(4S=4+4^2+4^3+...+4^{24}\)
\(S=\frac{4S-S}{3}=\frac{4^{24}-1}{3}\)
b/ Xem lại đề bài\(3S=4^{6x}-1=4^{24}-1\Rightarrow6x=24\Rightarrow x=4\)