Chủ đề này có 1 trả lời

[raquaza]

Chứng minh rằng $(C_{4026-k}^{2013})(C_{4026+k}^{2013})\leq (C_{4026}^{2013})^{2}$ $(2013\leq k\leq 4026)$ $k\in N$

[chanhquocnghiem]

Chứng minh rằng $(C_{4026-k}^{2013})(C_{4026+k}^{2013})\leq (C_{4026}^{2013})^{2}$ $(2013\leq k\leq 4026)$ $k\in N$

Sửa lại đề : $k\in N;0\leqslant k\leqslant 2013$

Ta có : 

$C_{4026-k}^{2013}=\frac{(4026-k)(4025-k)...(2014-k)}{2013!}$ (1)

$C_{4026+k}^{2013}=\frac{(4026+k)(4025+k)...(2014+k)}{2013!}$ (2)

---> $(C_{4026-k}^{2013})(C_{4026+k}^{2013})=\frac{(4026^2-k^2)(4025^2-k^2)...(2014^2-k^2)}{(2013!)^2}$ (3)

$(C_{4026}^{2013})^2=(\frac{4026.4025...2014}{2013!})^2=\frac{4026^2.4025^2...2014^2}{(2013!)^2}$ (4)

Từ (3),(4) ---> $(C_{4026-k}^{2013})(C_{4026+k}^{2013})\leqslant (C_{4026}^{2013})^2$ ($k\in N;0\leqslant k\leqslant 2013$)

Dấu bằng xảy ra khi $k=0$.