수열의 합
1차 수열의 합
\[ \sum_{i=1}^{n} i = \frac{n(n+1)}2\]
\[\begin{array}{llllllll} \underset {i=1}{ \overset {n}{\sum}}i + \underset {i=n}{ \overset {1}{\sum}} i &= 1 &+ 2
&+ 3 &+ ~ ... ~ &+ (n - 2) &+ (n - 1) &+ n \\
&~~+ n &+ (n - 1) &+ (n - 2) &+ ~ ... ~ &+ 3 &+ 2 &+ 1 \\
&= (n + 1) &+(n + 1) &+(n + 1) &+ ~ ... ~ &+(n + 1) &+(n + 1) &+(n + 1) \\
2 \underset {i=1}{ \overset {n}{\sum}}i &= n(n+1)
\end{array}\]
제곱 수열의 합
\[ \therefore \sum_{i=1}^{n} {i^2} = \frac{n(n+1)(2n+1)}{6} \]
- 다음 수열의 합을 나열하여 도출합니다.
\[\begin{align} \sum_{i=1}^{n} \{ i^3 - (i+1)^3 \} &= (1 -\cancel{2^3}) + (\cancel{2^3} - \cancel{3^3}) + ... +
(\cancel{n^3} - (n+1)^3)
\\ &= 1 - (n+1)^3
\\ &= \cancel 1 -n^3 -3n^2 -3n - \cancel 1
\end{align}\]
-
위 수열을 합을 다시 교환, 결합법칙을 이용하여 정리합니다.
\[\begin{aligned} \sum_{i=1}^{n} \{ i^3 - (i+1)^3 \} &= \sum_{i=1}^{n} \{ -3i^2 - 3i - 1 \}
\\ &= - 3 \sum_{i=1}^{n} {i^2} - 3 \sum_{i=1}^{n} i - \sum_{i=1}^{n} 1
\\ &= - 3 \sum_{i=1}^{n} {i^2} - 3 \frac{n(n+1)}{2} - n = -n^3 -3n^2 -3n
\end{aligned}\]
-
위 두 결론을 결합하여 정리합니다.
\[\begin{array}{lll} 3 \underset {i=1}{ \overset {n}{\sum}} {i^2} &+ 3 \displaystyle{\frac{n(n+1)}{2}} + n
&= n^3 +3n^2 +3n
\\ 3 \underset {i=1}{ \overset {n}{\sum}} {i^2} &+ 3 {n(n+1)} + 2n &= 2n^3 +6n^2 +6n
\\ 6 \underset {i=1}{ \overset {n}{\sum}} {i^2}& &= 2n^3 +6n^2 +6n -3n^2 - 3n - 2n
\\ & &= 2n^3 + 3n^2 + n
\\ & &= n(n+1)(2n+1)
\end{array}\]
3제곱 수열의 합
\[ \sum_{i=1}^{n} {i^3} = \left({\frac {x(x+1)}2}\right)^2 \]
\[\begin{aligned} \sum_{i=1}^{n} \{ (i+1)^4 - i^4 \} &= x^4 + 4x^3 + 6x^2 + 4x
\\ &= 4 \sum_{i=1}^{n} {i^3} + 6 \sum_{i=1}^{n} {i^2} + 4 \sum_{i=1}^{n} i + \sum_{i=1}^{n} 1
\\ &= 4 \sum_{i=1}^{n} {i^3} + n(n+1)(2n+1) + 2n(n+1) + n
\\ 4 \sum_{i=1}^{n} {i^3} &= x^4 + (4-2)x^3 + (6-3-2)x^2 + (4 -1 -2 -1)x
\\ &= x^4 + 2x^3 + x^2
\\ \sum_{i=1}^{n} {i^3} &= \left({\frac {x(x+1)}2}\right)^2
\end{aligned}\]