n(n+1) | ||
13+23+...+n3=(1+2+3+4+...+n)2=( | )2 | |
2 |
(k+1)(k+2) | ||
13+..+k3+(k+1)3=(1+2+..+k+k+1)2=( | )2 | |
2 |
(k+1)(k+2) | ||
13+..+k3+(k+1)3=( | )2 (ze wzoru na | |
2 |
(k+1)(k+2) | ||
s.wyr.ciąg.arytm.)=( | )2 | |
2 |
(k+1)(k+2) | ||
13+..+k3+(k+1)3=( | )2 | |
2 |
k(k+1) | ||
13 + ... + k3 + (k+1)3 = // z (2) czyli dla n=k // = ( | )2 + (k+1)3 = | |
2 |
1 | 1 | 1 | ||||
= | [ k2(k+1)2 + 4(k+1)3] = | (k+1)2[ k2 + 4(k+1)] = | (k+1)2[ k2 + 4k + | |||
4 | 4 | 4 |
1 | (k+1)(k+2) | |||
= | (k+1)2[k+2]2 = ( | )2 | ||
4 | 2 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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