Question

CMR: Với mọi n\(\in\)N,ta đều có:

\(\left[\frac{n+3}{4}\right]+\left[\frac{n+5}{4}\right]+\left[\frac{n}{2}\right]=n+1\)

Answer 1

Xét n trong các trường hợp sau:

+) n = 4k (k \(\in\) N) => VT = \(\left[\frac{4k+3}{4}\right]+\left[\frac{4k+5}{4}\right]+\left[\frac{4k}{2}\right]=\left[k+0,75\right]+\left[k+1,25\right]+\left[2k\right]\)

\(=k+\left(k+1\right)+2k=4k+1=n+1\)= VP

+) n = 4k + 1 (k \(\in\) N) => VT = \(\left[\frac{4k+4}{4}\right]+\left[\frac{4k+6}{4}\right]+\left[\frac{4k+1}{2}\right]=\left[k+1\right]+\left[k+1,5\right]+\left[2k+0,5\right]\)

\(=\left(k+1\right)+\left(k+1\right)+2k=4k+2=n+1\)= VP

+) n = 4k + 2 (k \(\in\) N)   => VT= \(\left[\frac{4k+5}{4}\right]+\left[\frac{4k+7}{4}\right]+\left[\frac{4k+2}{2}\right]=\left[k+1,25\right]+\left[k+1,75\right]+\left[2k+1\right]\)

\(=\left(k+1\right)+\left(k+1\right)+\left(2k+1\right)=4k+3=n+1\)= VP

+) n = 4k + 3 (k \(\in\) N)  => VT = \(\left[\frac{4k+6}{4}\right]+\left[\frac{4k+8}{4}\right]+\left[\frac{4k+3}{2}\right]=\left[k+1,5\right]+\left[k+2\right]+\left[2k+1,5\right]\)

\(=\left(k+1\right)+\left(k+2\right)+\left(2k+1\right)=4k+4=n+1\)= VP

Từ các trường hợp trên => đpcm

 

Answer 2

\(\frac{n+3}{4}+\frac{n+5}{4}+\frac{n}{2}=\frac{n+3}{4}+\frac{n+5}{4}+\frac{2n}{4}=\frac{n+3+n+5+2n}{4}=\frac{4n+8}{4}=n+2\)