obliczyc warosc sumy
: 29 sty 2013, o 20:23
Obliczyc
\(\displaystyle{ \sum_{k=0}^{n}(k+1)5^k {n \choose k}}\)
Moje rozwiazanie:
\(\displaystyle{ \sum_{k=0}^{n}(k+1)5^k {n \choose k}=\sum_{k=0}^{n}k5^k {n \choose k}+\sum_{k=0}^{n}5^k {n \choose k}=\left[ \sum_{k=0}^{n}5^k {n \choose k}\right] '+\sum_{k=0}^{n}5^k {n \choose k}=\left[ \sum_{k=0}^{n} {n \choose k}x^k-5^{n-k}\right]' + \sum_{k=0}^{n} {n \choose k}x^k-5^{n-k} =n6^{n-1}+6^n}\)
Tak powinno wyjsc?
\(\displaystyle{ \sum_{k=0}^{n}(k+1)5^k {n \choose k}}\)
Moje rozwiazanie:
\(\displaystyle{ \sum_{k=0}^{n}(k+1)5^k {n \choose k}=\sum_{k=0}^{n}k5^k {n \choose k}+\sum_{k=0}^{n}5^k {n \choose k}=\left[ \sum_{k=0}^{n}5^k {n \choose k}\right] '+\sum_{k=0}^{n}5^k {n \choose k}=\left[ \sum_{k=0}^{n} {n \choose k}x^k-5^{n-k}\right]' + \sum_{k=0}^{n} {n \choose k}x^k-5^{n-k} =n6^{n-1}+6^n}\)
Tak powinno wyjsc?