Question

求下列各式中的n值:

(1)

(2)

(3)

Answer 1

解:(1)由排列數(shù)公式、得

(2n+1)·2n·(2n-1)·(2n-2)=140·n(n-1)(n-2),

整理得4n²-35n+69=0,∴(4n-23)(n-3)=0.

∴n=3或n=(舍去).∴n=3.

(2)由排列數(shù)公式,得3n(n-1)(n-2)=2(n+1)·n+6n(n-1),

整理得3n²-17n+10=0,解得n=5或n=(舍去),∴n=5.

(3)由排列數(shù)公式、得

化簡(jiǎn),得n²-19n+78=0.n=6或n=13.

∵n≤8,∴n=6.