Question

1*n^2+2*(n-1)^2+3*(n-2)^2+……+n*1^2=？

只要一个答案就可以

Answer 1

ak=k(n+1－k)²=k(n+1)²－2k²(n+1)+k³
Sn∑ak=n(n+1)³/2－n(n+1)²(2n+1)/3+n²(n+1)²/4
=n(n+1)²/12*(6(n+1)－4(2n+1)+3n)
=n(n+1)²(n+2)/12

Answer 2

ak
= k^2.(n-k+1)^2
= k^2.[ (n+1)^2 - 2(n+1)k + k^2 ]
=(n+1)^2. k^2 - 2(n+1)k^3 + k^4
1*n^2+2*(n-1)^2+3*(n-2)^2+……+n*1^2
=a1+a2+...+an
=(n+1)^2 . [(1/6)n(n+1)(2n+1)] -2(n+1).{(1/4)[n(n+1)]^2} +(1/6)n(n+1)^2.(n^2+2)
=(1/6)n(n+1)^3.(2n+1) -(1/2)n(n+1)^3 +(1/30)n(n+1)(6n^3+9n^2 +11n+4 )
//

bk= k^2
Bn = b1+b2+...+bn = (1/6)n(n+1)(2n+1)
ck
= k^3
= (k-1)k(k+1) +k
=(1/4)[ (k-1)k(k+1)(k+2) -(k-2)(k-1)k(k+1)] + (1/2)[ k(k+1) -(k-1)k]
Cn
=c1+c2+..+cn
=(1/4)(n-1)n(n+1)(n+2) + (1/2)n(n+1)
=(1/4)n(n+1) [ (n-1)(n+2) + 2]
=(1/4)n(n+1)(n^2+n)
=(1/4)[n(n+1)]^2
dk

= k^4
=(k-1)k(k+1)(k+2) - 2k^3 + k^2 +2k
=(1/5)[(k-1)k(k+1)(k+2)(k+3)-(k-2)(k-1)k(k+1)(k+2)] - 2k^3 + k^2 +2k

Dn
=d1+d2+...+dn
=(1/5)(n-1)n(n+1)(n+2)(n+3) -2{(1/4)[n(n+1)]^2}
+2{ (1/6)n(n+1)(2n+1) } + 2[ n(n+1)/2]
=(1/5)(n-1)n(n+1)(n+2)(n+3) -(1/2)[n(n+1)]^2 +(1/3)n(n+1)(2n+1) + n(n+1)
=(1/30)n(n+1) [ 6(n-1)(n+2)(n+3) - 15n(n+1) +10(2n+1) +30 ]
=(1/30)n(n+1) [ 6n^3+24n^2 +6n-36 -15n^2-15n +20n+10 +30]
=(1/30)n(n+1)(6n^3+9n^2 +11n+4 )

Answer 3

1*n^2+2*(n-1)^2+3*(n-2)^2+……+n*1^2

=  n (n + 1)^2 (n + 2)/12