Question

求数列前n项和:1²+3²+5²+……+(2n-1)²=? 2²+4²+……+（2n）²=?

Answer 1

解答：
需要用到平方和公式
1²+2²+3²+......+n²=n(n+1)(2n+1)/6
∵ (2n-1)²=4n²-4n+1
则 1²+3²+5²+……+(2n-1)²
= (4*1²-4+1)+(4*2²-4*2+1)+(4*3²-4*3+1)+........+(4n²-4n+1)
=4*(1²+2²+3²+......+n²)-4(1+2+3+......+n)+n
=4n(n+1)(2n+1)/6-4(n+1)*n/2+n
=2n(n+1)(2n+1)/3-2n(n+1)+n
=(4n³)/3-n/3

∵ (2n)²=4n²
2²+4²+……+（2n）²
=4(1²+2²+3²+......+n²)
=2n(n+1)(2n+1)/3

Answer 2

因为(2n-1)²=4*n^2-4n+1

所以，1²+3²+5²+……+(2n-1)²
=4*（1^2+2^2+...+n^2)-4*(1+2+...+n)+(1+1+...+1)
=4*n(n+1)(2n+1)/6-4*n(n+1)/2+n
=n(n+1)*((4n+2)/3-2)+n
=4n(n+1)(n-1)/3+n

因为（2n）²=4*n^2
所以，
2²+4²+……+（2n）²
=4*（1^2+2^2+...+n^2)
=4*n(n+1)(2n-+)/6
=2n(n+1)(2n+1)/3

Answer 3

可以设Tn=1²+3²+5²+……+(2n-1)²=an³+bn²+cn+d
则T1=a+b+c+d=1
T2=8a+4b+2c+d=10
T3=27a+9b+3c+d=35
T4=6a+16b+4c+d=84，可以解出啊a、b、c、d
同理可以得到2²+4²+……+（2n）²=求和公式

Answer 4

1.1²+3²+5²+……+(2n-1)²=sn
sn=(1+(2n-1))^2-2*(2n-1)+(3+(2n-3))^2-2*3(2n-3)......+（（n-2)+(n+2))^2-2(n-2)(n+2)+n^2=2n^3-2(1*(2n-1）+3（2n-3)+.......+(n-2)(2n-(n-2)))=2n^3-2n^2-n+2
下一题的方法同理，自己整理吧

Answer 5

解：
∵
1²+2²+3²+...+n²=n(n+1)(2n+1)/6
因此：
1²+2²+3²+...+(2n)² =2n(2n+1)(4n+1)/6=n(2n+1)(4n+1)/3
又∵
[1²+3²+5²+……+(2n-1)²]+[2²+4²+……+(2n)²] = 1²+2²+3²+...+(2n)²
而：
[2²+4²+……+(2n)²] - [1²+3²+5²+……+(2n-1)²]
=(2-1)(2+1)+(4-3)(4+3)+.......+(2n-2n+1)(2n+2n-1)
=3+7+11+15+....+4n-1
=n(2n+1)
∴
2 [1²+3²+5²+……+(2n-1)²]+n(2n+1) = n(2n+1)(4n+1)/3
∴
1²+3²+5²+……+(2n-1)² = (1/2)[n(2n+1)(4n+1)/3 - n(2n+1)] = n(2n+1)(2n-1)/3
2²+4²+……+(2n)² = n(2n+1)(2n-1)/3 +n(2n+1) = n(2n+1)(2n+2)/3