Question

数列题求解：

已知连续2n+1个正整数总和为a，且这些数中后n个数的平方和与前n个数的平方和之差为 b．若a/b=11/60，则n的值为______________

Answer 1

解：设中间的数，即第n+1个数为m，
则连续的数 为 m-n,m-n+1,......，m-1，m，m+1，。。。m+n-1，m+n
a=Sn=m*（2n+1）
b=（m+n)^2- (m-n)^2 +[m+(n-1)]^2-[m-(n-1)]^2+[m+(n-2)]^2-[m-(n-2)]^2 +.....+(m+1)^2 -(m-1)^2
= 4mn +4m(n-1)+4m(n-2)+....+4m*2+4m
=4mn*(n+1)/2
=2mn(n+1)
因为a/b=11/60
m*(2n+1)/[2mn(n+1)] =11/60
(2n+1)/[2n(n+1)] =11/60
22n^2+22n =120n+60
22n^2-98n-60=0
11n^2-49n-30=0
(n-5)(11n+6) =0
解得 n=5，或者n=-6/11（舍去）
所以n=5

Answer 2

x+(x+1)+(x+2)+...+(x+2n)=(2n+1)x+n(2n+1)=a = (x+n)(2n+1)
(x+2n)^2+(x+2n-1)^2+...+(x+2n-n+1)^2 - [x^2+(x+1)^2+...+(x+n-1)^2] = b
= [(x+2n)^2-(x+n-1)^2] + [(x+2n-1)^2-(x+n-2)^2] + ... + [(x+2n-n+1)^2-(x+n-1-n+1)^2]
= (2x+3n-1)(n+1) + (2x+3n-1-2)(n+1) + ... + (2x+3n-1-2n+2)(n+1)
=(n+1)[(2x+3n-1)n - (2+4+...+2(n-1))]
=(n+1)[(2x+3n-1)n - n(n-1)]
=n(n+1)[2x+3n-1-n+1]
=n(n+1)(2x+2n)
=2n(n+1)(x+n),

11/60 = a/b = (x+n)(2n+1)/[2n(n+1)(x+n)] = (2n+1)/[2n(n+1)],

22n(n+1) = 60(2n+1), 11n(n+1) = 30(2n+1) = 60n + 30 = 11n^2 + 11n,
0=11n^2 - 49n - 30 = (11n+6)(n-5),
n=5