wzory na sume
jedrzej: Witam! Prosze o pomoc w wyjasnieniu wyciagania wzorw na pewne sumy ciagow:
np jak oblcizyc ze:
| | n(n+1)(2n+1) | |
12 + 22....n2 = |
| |
| | 6 | |
| | n(2n−1)(2n+1) | |
albo ze 12 +32 + 52...(2n−1)2 = |
| |
| | 3 | |
istnieje jakas metoda albo co?
Mila:
Skorzystamy z wzoru
1) (n+1)
3=n
3+3n
2+3n+1
2) (n+1)
3−n
3=n
3+3n
2+3n+1−n
3=3n
2+3n+1
S
n=0
3+1
3+2
3+3
3+4
3+5
3+.............+(n−1)
3+n
3
S
n+1=1
3+2
3+3
3+4
3+5
3+.............+(n−1)
3+n
3+(n+1)
3
S=S
n+1−S
n=
(n+1)3
Obliczymy sumę S inaczej, odejmując a
n+1−a
n
S:
(n+1)
3−n
3+(n
3−(n−1)
3+................+(3
3−2
3)+(2
3−1
3)+(1
3−0
3)=
=(3n
2+3n+1)+................(3*3
2+3*3+1)+(3*2
2−3*2+1)+(3*1
2−3*1+1)=
=(3*n
2+3*(n−1)
2+..........3
2+2
2+1
2)+(3*n+3*(n−1)+3(n−2)+......3+2+1)+(n+1)*1=
| | n*(n+1) | |
S=3*(12+22+32+.....+n2)+3* |
| +n+1 |
| | 2 | |
| | 3n2+5n+2 | |
S=3*(12+22+32+.....+n2)+ |
| ⇔ |
| | 2 | |
| | 3n2+5n+2 | |
(n+1)3=3*(12+22+32+.....+n2)+ |
| |
| | 2 | |
| | 2*(n+1)3−3n2−5n−2 | |
3*(12+22+32+.....+n2)= |
| |
| | 2 | |
| | 1 | | 2n3+3n+n | |
(12+22+32+.....+n2)= |
| * |
| ⇔ |
| | 3 | | 2 | |
| | 1 | | | |
(12+22+32+.....+n2)= |
| * |
| |
| | 3 | | 2 | |
| | n*(n+1)*(2n*1 | |
(12+22+32+.....+n2)= |
| |
| | 6 | |
===================================