Bo ja tu nie widzę rozwiązania, tylko jego próby
| n(n+1)(2n+1) | ||
∑(n+1) k2 = (n+1)2 + ∑n k2 = // z (2) // = (n+1)2 + | = | |
| 6 |
| 6(n+1)2 + n(n+1)(2n+1) | (n+1)[ 6n + 6 + n(2n+1)] | |||
= | = | = | ||
| 6 | 6 |
| (n+1)[ 4n + 2 + 2n + 4 + n(2n+1)] | (n+1)[ 2n + 4 + (n+2)(2n+1)] | |||
= | = | = | ||
| 6 | 6 |
| (n+1)[ 2(n + 2) + (n+2)(2n+1)] | (n+1)(n+2)(2n+1+2) | |||
= | = | = | ||
| 6 | 6 |
| (n+1)(n+2)(2n+3 | ||
= | = P | |
| 6 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
|---|---|
| Twój nick | |
