Since \( n = \sum_{i=1}^{n} 1 \) we can write that \( n = \sum_{i=1}^{n} i - ( i - 1 ) \)
\( n = \sum_{i=1}^{n} i - ( i - 1 ) \)
\( n = \sum_{i=1}^{n} 1 \)
\( n = S_n(0) \)
\( S_n(0) = n \)
\( n^2 = \sum_{i=1}^{n} i^2 - ( i - 1 )^2 \)
\( n^2 = \sum_{i=1}^{n} i^2 - ( i^2 - 2 i + 1 \)
\( n^2 = \sum_{i=1}^{n} i^2 - i^2 + 2 i - 1 \)
\( n^2 = \sum_{i=1}^{n} 2 i - 1 \)
\( n^2 = 2 S_n(1) - S_n(0) \)
\( - 2 S_n(1) = - n^2 - S_n(0) \)
\( 2 S_n(1) = n^2 + n \)
\( S_n(1) = \frac{ n^2 + n }{2} \)
\( S_n(1) = \frac{(n)(n+1)}{2} \)
\( n^3 = \sum_{i=1}^{n} i^3 - ( i - 1 )^3 \)
\( n^3 = \sum_{i=1}^{n} i^3 - ( i^3 - 3 i^2 + 3 i - 1 ) \)
\( n^3 = \sum_{i=1}^{n} i^3 - i^3 + 3 i^2 - 3 i + 1 \)
\( n^3 = \sum_{i=1}^{n} 3 i^2 - 3 i + 1 \)
\( n^3 = 3 S_n(2) - 3 S_n(1) + S_n(0) \)
\( -3 S_n(2) = - n^3 - 3 \left ( \frac{(n)(n+1)}{2} \right ) + ( n ) \)
\( 3 S_n(2) = n^3 + 3 \left ( \frac{(n)(n+1)}{2} \right ) - ( n ) \)
\( 3 S_n(2) = n^3 + \frac{(3)(n)(n+1)}{2} - n \)
\( S_n(2) = \frac{ n^3}{3} + \frac{(n)(n+1)}{2} - \frac{n}{3} \)
\( S_n(2) = \frac{ n^3}{3} + \frac{n^2+n}{2} - \frac{n}{3} \)
\( S_n(2) = \frac{ 2 n^3}{6} + \frac{3 n^2+3 n}{6} - \frac{2 n}{6} \)
\( S_n(2) = \frac{ 2 n^3 + 3 n^2 + 3 n - 2 n}{6} \)
\( S_n(2) = \frac{ 2 n^3 + 3 n^2 + n}{6} \)
\( n^4 = \sum_{i=1}^{n} i^4 - ( i - 1 )^4 \)
\( n^4 = \sum_{i=1}^{n} i^4 - ( i^4 - 4 i^3 + 6 i^2-4 i+1) \)
\( n^4 = \sum_{i=1}^{n} i^4 - i^4 + 4 i^3 - 6 i^2 + 4 i - 1 \)
\( n^4 = \sum_{i=1}^{n} 4 i^3 - 6 i^2 + 4 i - 1 \)
\( n^4 = 4 S_n(3) - 6 S_n(2) + 4 S_n(1) - S_n(0) \)
\( - 4 S_n(3) = - n^4 - 6 \left ( \frac{ 2 n^3 + 3 n^2 + n}{6} \right ) + 4 \left ( \frac{(n)(n+1)}{2} \right ) - ( n )\)
\( - 4 S_n(3) = - n^4 - ( 2 n^3 + 3 n^2 + n ) + 2( (n)(n+1) ) - n \)
\( - 4 S_n(3) = - n^4 - 2 n^3 - 3 n^2 - n + 2( n^2 + n ) - n \)
\( - 4 S_n(3) = - n^4 - 2 n^3 - 3 n^2 - n + 2 n^2 + 2 n - n \)
\( 4 S_n(3) = n^4 + 2 n^3 + 3 n^2 + n - 2 n^2 - 2 n + n \)
\( 4 S_n(3) = n^4 + 2 n^3 + n^2 \)
\( 4 S_n(3) = n^2 (n^2 + 2 n + 1) \)
\( 4 S_n(3) = n^2 (n+1)^2 \)
\( S_n(3) = \frac{ n^2 (n+1)^2}{4} \)