tính: (1+2+3+...+100) . (1 mu 2+ 2 mu 2+3 mu 2+...+100 mu 2) . (65.111-13.15.37)
Những câu hỏi liên quan
Tính tổng
B=3-3 mu 2 +3 mu 3 -......- 3 mu 100
A=1+2+ 2 mu 2 +......+2 mu 100
\(A=1+2+2^2+...+2^{100}\)
\(2A=2+2^2+2^3+..,+2^{101}\)
\(2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)
\(A=2^{101}-1\)
\(B=\)\(3-3^2+3^3-...-3^{100}\)
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tính tổng
B= 7-7 mu 4 + 7 mu 4 -........+7 mu 301
A = 1 + 5 mũ 2 + 5 mu 4 + 5 mu 6 +.....+5 mu 200
tính
A= 1/7+1/7mu 2 + 1/7 mu 3+......+1/7mu 100
B=-4/5+4/5 mu 2 - 4/5 mu 3 + ....+4/5mu 200
tính A=25 mũ 8 + 25 mũ 4 + 25 mu 20 +......+25 mu 4 +1 / 25 mu 20 + 25 mu 28 + 25 mu 26 +.....= 25 mu 2 +1
7 × 3 mu x + 20 × 3 mu x = 3 mu 25
tinh tong s=1 mu 2 cong 2 mu 2 conng 3 mu 2 cong .... cong 100 mu 2
Xem chi tiết
Ta có :\(n^2-n=n.\left(n-1\right)\)
\(\implies\) \(n^2=\left(n-1\right)n+n\)
Áp dụng : với n=2017 thay vào ta có:
\(S=1+1.2+2+2.3+3+...+99.100+100\)
\(S=\left(1.2+2.3+...+99.100\right)+\left(1+2+...+100\right)\)
\(S=\frac{99.100.101}{3}+\frac{101.100}{2}\)
\(S=100.101.\left(\frac{99}{3}+\frac{1}{2}\right)\)
\(S=\frac{100.101.201}{6}\)
Tinh tong sau hop li :
A=2+2 mu 2 + 2 mu 3 + ............+2 mu 100
B=1+3 mu 2 + 3 mu 3 + .......... + 3 mu 2009
C = 4+ 4 mu 2 + 4 mu 3 + ...............+ 4 mu n
Giup minh voi sory may minh ko co dau minh dang can gap lam !!!
Lần sau viết cái đề rõ rõ ra nhs!!!
a) \(A=2+2^2+2^3+................+2^{100}\)
\(\Rightarrow2A=2^2+2^3+2^4+................+2^{100}+2^{101}\)
\(\Rightarrow2A-A=\left(2^2+2^3+..............+2^{100}+2^{101}\right)-\left(2+2^2+............+2^{100}\right)\)
\(\Rightarrow A=2^{101}-2\)
b) \(B=1+3+3^2+..................+3^{2009}\)
\(\Rightarrow3B=3+3^2+3^3+..................+3^{2009}+3^{2010}\)
\(\Rightarrow3B-B=\left(3+3^2+...............+3^{2010}\right)-\left(1+3+3^2+.............+3^{2009}\right)\)
\(\Rightarrow2B=3^{2010}-1\)
\(\Rightarrow B=\dfrac{3^{2010}-1}{2}\)
c) \(C=4+4^2+4^3+................+4^n\)
\(\Rightarrow4C=4^2+4^3+.................+4^n+4^{n+1}\)
\(\Rightarrow4C-C=\left(4^2+4^3+.............+4^n+4^{n+1}\right)-\left(4+4^2+............+4^n\right)\)
\(\Rightarrow3C=4^{n+1}-4\)
\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)
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\(A=2+2^2+2^3+...+2^{100}\)
\(2A=2^2+2^3+2^4+...+2^{101}\)
\(\Rightarrow2A-A=2^{101}-2\)
\(\Rightarrow A=2^{101}-2\)
Vậy \(A=2^{101}-2\).
\(B=1+3^2+3^3+...+3^{2009}\)
\(3B=3+3^3+3^4+...+3^{2010}\)
\(\Rightarrow3B-B=3^{2010}-7\)
\(\Rightarrow2B=3^{2010}-7\)
\(\Rightarrow B=\dfrac{3^{2010}-7}{2}\)
Vậy \(B=\dfrac{3^{2010}-7}{2}\).
\(C=4+4^2+4^3+...+4^n\)
\(4C=4^2+4^3+4^4+...+4^{n+1}\)
\(\Rightarrow4C-C=4^{n+1}-4\)
\(\Rightarrow3C=4^{n+1}-4\)
\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)
Vậy \(C=\dfrac{4^{n+1}-4}{3}\).
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chung minh rang 1\2 mu 2+1\3 mu 2+1\4 mu2+...+1\100 mu 2 < 1
ai lam day du dau tien minh se k cho nha
minh can gap lam
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\).
A=3 mu 100 - 3 mu 99 +3 mu 98 - 3 mu 97 + ... +3 mu 2 - 3 mu 1. Tim n biet 3 + 4A = 3 mu n
Cho a = 2 + 2 mu 2 + 2 mu 3 +.....+ 2 mu 100
tính 2.a
tìm x biết
2.a -a=2 mu x -2
\(2a=2.\left(2+2^2+...+2^{100}\right)\)
\(2a=2^2+2^3+...+2^{101}\)
\(2a-a=\left(2^2+2^3+...+2^{101}\right)-\left(2^1+2^2+...+2^{100}\right)\)
\(2a-a=2^{101}-2\)
mà \(2a-a=2^x-2=>x=101\)
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Xem thêm câu trả lời
chung to H chia het cho 155 , biet H = 2 + 2 mu 2 + 2 mu 3 + 2 mu 4 + ......+ 2 mu 99 + 2 mu 100
so sanh
(-1/16) mu 100 va (-1/2) mu 500
(1/81) mu 12 va (1/27) mu 16
(-2) mu 10 va 1000
2 mu 93 va 5 mu 35
a: \(\left(-\dfrac{1}{16}\right)^{100}=\left(\dfrac{1}{16}\right)^{100}=\left(-\dfrac{1}{2}\right)^{400}\)
\(\left(-\dfrac{1}{2}\right)^{500}=\left(-\dfrac{1}{2}\right)^{500}\)
mà \(400< 500\)
nên \(\left(-\dfrac{1}{16}\right)^{100}< \left(-\dfrac{1}{2}\right)^{500}\)
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