S = 1 + 2q + 3q2 + ....+ (n+1)qn
q.S = q + 2q2 + 3q3 + ...+ (n+1)qn+1
=> S - q.S = 1 + q + q2 + q3 + ...+ qn - (n+1)qn+1
=> (1 - q).S = (1 + q + q2 + q3 + ...+ qn) - (n+1)qn+1
Tính A = 1 + q + q2 + q3 + ...+ qn => q.A = q + q2 + ...+ qn+ 1
=> A - q.A = 1 - qn+1 => A = (1 - qn+1)/(1-q)
Vậy (1-q).S = (1 - qn+1)/(1-q) - (n+1)qn+1 => S = (1 - qn+1)/(1-q)2 - (n+1)qn+1/ (1 - q)
\(A=1.2+2.3+...+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.3+2.3.3+...+n\left(n+1\right)3\)
\(=1.2.3+2.3.\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
cho Sn= \(\frac{1}{1\cdot2\cdot3\cdot4}\)+ \(\frac{1}{2\cdot3\cdot4\cdot5}\)+ ... + \(\frac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)}\)
CMR: 18<\(\frac{1}{S_n}\)<=24
Bài 1 :a, Tính tổng\(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+.......+\left(-\frac{1}{7}\right)^{2007}\)
b, CMR \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.......+\frac{99}{100!}<1\)
c, CMR: mọi số nguyên dương n thì: \(3^{n+2}-2^{n+2}+3^n-2^n\)chia hết cho 10
Tính :
1 + 3 + 5 + 7 + ... + (2n - 1) = 225
Giải :
Theo công thức tính dãy số , ta có :
\(\frac{\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}.\left[\left(2n-1\right)+1\right]}{2}=225\)
\(\frac{\left\{\left[2n-2\right]:2+1\right\}.2n}{2}=225\)
\(\left\{\left[2n-2\right]:2+1\right\}.n=450\)(Lượt giản thừa số 2)
\(\left\{\frac{2n-2}{2}+1\right\}.n=225\)
\(\left\{\frac{2n-2}{2}+\frac{2}{2}\right\}.n=225\)
\(\frac{2n-2+2}{2}.n=225\)
\(\frac{2n}{2}.n=225\)
\(n^2=225\)
\(\Rightarrow n=\sqrt{225}=15\)
Giúp mik với
Tính nhanh:
a. A=\(\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\left(n\in N\right)\)
b. B=\(\left(10000-1^2\right)\left(10000-2^2\right)\left(10000-3^2\right)..\left(10000-1000^2\right)\)
c. C=\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
d. D=\(1999^{\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-10^3\right)}\)
Bài 1: Tính các biểu thức:
A= \(\left(\frac{3}{4}\right)^{-4}.\left(-\frac{2}{3}\right)^{-3}\)
B= \(\left(4^3\right)^{-2}\). \(a^{2015}\)
C= \(\left[\left(-\frac{1}{3}\right).\frac{2}{5}.\left(-\frac{3}{4}\right)\right]^3\)
Bài 2: CMR: \(\forall\) n\(\in\)Z
\(\left(3^{n+2}-2^{n+2}+3^n-2^n\right)⋮10\)
tính S1
\(S_1=\frac{1}{1\cdot2\cdot3\cdot4\cdot5}\)\(+\frac{1}{2\cdot3\cdot4\cdot5\cdot6}+.................+\frac{1}{\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(2\right)}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\left(n\in N\right)\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+.......+\frac{1}{20}\left(1+2+3+4....+20\right)\)
tính các tích sau với nEN, n lớn hơn bằng 2
a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)
b)\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)
c)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
