Let
S_1 =\sum\limits_{k=1}^n1 = 1+1+1+...+1 = 1\cdot n = n S 1 = k = 1 ∑ n 1 = 1 + 1 + 1 + ... + 1 = 1 ⋅ n = n S_2 =\sum\limits_{k=1}^nk=\frac{n(n+1)}{2} S 2 = k = 1 ∑ n k = 2 n ( n + 1 ) S_3 = \sum\limits_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6} S 3 = k = 1 ∑ n k 2 = 6 n ( n + 1 ) ( 2 n + 1 ) S_4 = \sum\limits_{k=1}^nk^3 S 4 = k = 1 ∑ n k 3
Use the fact
(k+1)^4 -k^4 =4k^3 +6k^2 +4k+1 ( k + 1 ) 4 − k 4 = 4 k 3 + 6 k 2 + 4 k + 1
Put k=1,2,3..., n k = 1 , 2 , 3... , n to this formula:
2^4 -1^4 =4\cdot 1^3+6\cdot 1^2 +4\cdot 1+1 2 4 − 1 4 = 4 ⋅ 1 3 + 6 ⋅ 1 2 + 4 ⋅ 1 + 1
3^4 -2^4 =4\cdot 2^3+6\cdot 2^2 +4\cdot 2+1 3 4 − 2 4 = 4 ⋅ 2 3 + 6 ⋅ 2 2 + 4 ⋅ 2 + 1
4^4 -3^4 =4\cdot 3^3+6\cdot 3^2 +4\cdot 3+1 4 4 − 3 4 = 4 ⋅ 3 3 + 6 ⋅ 3 2 + 4 ⋅ 3 + 1
...
(n+1)^4 -n^4 =4\cdot n^3+6\cdot n^2 +4\cdot n+1 ( n + 1 ) 4 − n 4 = 4 ⋅ n 3 + 6 ⋅ n 2 + 4 ⋅ n + 1
Sum left sides of these equalities:
\cancel{2^4} -1^4 +\cancel{3^4} -\cancel{2^4} +\cancel{4^4} -\cancel{3^4} +...+(n+1)^4 -\cancel{n^4}=(n+1)^4-1 2 4 − 1 4 + 3 4 − 2 4 + 4 4 − 3 4 + ... + ( n + 1 ) 4 − n 4 = ( n + 1 ) 4 − 1
Sum right sides of these equalities:
4(1^3+2^3+...+n^3)+6(1^2+2^2+...+n^2)+4(1+2+...+n)+(1+1+...+1)= 4 ( 1 3 + 2 3 + ... + n 3 ) + 6 ( 1 2 + 2 2 + ... + n 2 ) + 4 ( 1 + 2 + ... + n ) + ( 1 + 1 + ... + 1 ) =
=4S_4+6S_3 +4S_2 +S_1 = 4 S 4 + 6 S 3 + 4 S 2 + S 1
Hence
4S_4+6S_3 +4S_2 +S_1 =(n+1)^4-1 4 S 4 + 6 S 3 + 4 S 2 + S 1 = ( n + 1 ) 4 − 1
4S_4=(n+1)^4-1-6S_3-4S_2-S_1 4 S 4 = ( n + 1 ) 4 − 1 − 6 S 3 − 4 S 2 − S 1
4S_4 =(n+1)^4 -1-\frac{6n(n+1)(2n+1)}{6}-\frac{4n(n+1)}{2}-n 4 S 4 = ( n + 1 ) 4 − 1 − 6 6 n ( n + 1 ) ( 2 n + 1 ) − 2 4 n ( n + 1 ) − n
4S_4 =(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n 4 S 4 = ( n + 1 ) 4 − 1 − n ( n + 1 ) ( 2 n + 1 ) − 2 n ( n + 1 ) − n
4S_4 =(n+1)^4-n(n+1)(2n+1)-2n(n+1)-(n+1) 4 S 4 = ( n + 1 ) 4 − n ( n + 1 ) ( 2 n + 1 ) − 2 n ( n + 1 ) − ( n + 1 )
4S_4 =(n+1)((n+1)^3 -n(2n+1)-2n-1) 4 S 4 = ( n + 1 ) (( n + 1 ) 3 − n ( 2 n + 1 ) − 2 n − 1 )
4S_4 =(n+1)(n^3+3n^2+\cancel{3n}+\cancel{1}-2n^2-\cancel{n}-\cancel{2n}-\cancel{1}) 4 S 4 = ( n + 1 ) ( n 3 + 3 n 2 + 3 n + 1 − 2 n 2 − n − 2 n − 1 )
4S_4 =(n+1)(n^3+n^2) 4 S 4 = ( n + 1 ) ( n 3 + n 2 )
4S_4 =(n+1)\cdot n^2(n+1) 4 S 4 = ( n + 1 ) ⋅ n 2 ( n + 1 )
4S_4 =n^2(n+1)^2 4 S 4 = n 2 ( n + 1 ) 2
S_4 =\frac{n^2(n+1)^2}{4}=(\frac{n(n+1)}{2})^2 S 4 = 4 n 2 ( n + 1 ) 2 = ( 2 n ( n + 1 ) ) 2
\sum\limits_{k=1}^nk^3 =(\frac{n(n+1)}{2})^2 k = 1 ∑ n k 3 = ( 2 n ( n + 1 ) ) 2