Chứng minh \(\sqrt{1+2+3+4+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}=n\)
giúp mik với ạ.
chứng minh rằng: \(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n+1\right)+}...+3+2+1=n\) với n∈N
Chứng minh rằng:\(\sqrt{1+2+3+...+\left(n+1\right)+n+\left(n-1\right)+...+3+2+1}\)=n
tính A = \(\left[\sqrt{2}\right]+\left[\sqrt[3]{\frac{3}{2}}\right]+\left[\sqrt[4]{\frac{4}{3}}\right]+...+\left[\sqrt[n+1]{\frac{n+1}{n}}\right]\)
CHỨNG MINH M
\(\sqrt{1+2+3+.....+\left(n-1\right)+n+\left(n-1\right)+.....+3+2+1}\)= n
giải hẳn ra
Chứng tỏ rằng : \(\text{A}=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]\).
Chứng minh rằng\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\) = n
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=\(1999^{\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-10^3\right)}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\left(n\in N\right)\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+.......+\frac{1}{20}\left(1+2+3+4....+20\right)\)