Oblicz wartość sumy
Otrzymałeś(aś) rozwiązanie do zamieszczonego zadania? - podziękuj autorowi rozwiązania! Kliknij
-
octahedron
- Expert
- Posty: 6762
- Rejestracja: 19 mar 2011, 00:22
- Otrzymane podziękowania: 3034 razy
- Płeć:
Re: Oblicz wartość sumy
\(\sum_{k=0}^{n}k\cdot 4^k=1\cdot 4^1+2\cdot 4^2+3\cdot 4^3+...+n\cdot 4^n=
=(4^1+4^2+4^3+...+4^n)+(4^2+4^3+...+4^n)+(4^3+...+4^n)+...+(4^{n-1}+4^n)+4^n=
=\frac{4(1-4^{n})}{1-4}+\frac{4^2(1-4^{n-1})}{1-4}+\frac{4^3(1-4^{n-2})}{1-4}+...+\frac{4^{n-1}(1-4^{2})}{1-4}+\frac{4^n(1-4)}{1-4}=
=\frac{4^{n+1}-4}{3}+\frac{4^{n+1}-4^2}{3}+\frac{4^{n+1}-4^3}{3}+...+\frac{4^{n+1}-4^{n-1}}{3}+\frac{4^{n+1}-4^n}{3}=
=\frac{1}{3}(n\cdot 4^{n+1}-4-4^2-4^3-...-4^n)=\frac{1}{3}\(n\cdot 4^{n+1}-\frac{4(1-4^n)}{1-4}\)=
=\frac{1}{3}\(n\cdot 4^{n+1}-\frac{4^{n+1}-4}{3}\)=\frac{(3n-1)4^{n+1}+4}{9}\)
lub
\(f(x)=\sum_{k=0}^{n}k\cdot x^k=x\sum_{k=1}^{n}k\cdot x^{k-1}=x\sum_{k=1}^{n}\(x^{k}\)'=x\(\sum_{k=1}^{n}x^{k}\)'=x\(x\cdot\frac{1-x^n}{1-x}\)'=x\(\frac{x-x^{n+1}}{1-x}\)'=\\=x\cdot\frac{(1-(n+1)x^{n})(1-x)-(x-x^{n+1})\cdot(-1)}{(1-x)^2}=x\cdot\frac{1-x-(n+1)x^{n}+(n+1)x^{n+1}+x-x^{n+1}}{(1-x)^2}=\\=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(1-x)^2}
f(4)=\sum_{k=0}^{n}k\cdot 4^k=\frac{n\cdot 4^{n+2}-(n+1)\cdot 4^{n+1}+4}{(1-4)^2}=\frac{4n\cdot 4^{n+1}-(n+1)\cdot 4^{n+1}+4}{9}=\frac{(3n-1)\cdot 4^{n+1}+4}{9}\)
=(4^1+4^2+4^3+...+4^n)+(4^2+4^3+...+4^n)+(4^3+...+4^n)+...+(4^{n-1}+4^n)+4^n=
=\frac{4(1-4^{n})}{1-4}+\frac{4^2(1-4^{n-1})}{1-4}+\frac{4^3(1-4^{n-2})}{1-4}+...+\frac{4^{n-1}(1-4^{2})}{1-4}+\frac{4^n(1-4)}{1-4}=
=\frac{4^{n+1}-4}{3}+\frac{4^{n+1}-4^2}{3}+\frac{4^{n+1}-4^3}{3}+...+\frac{4^{n+1}-4^{n-1}}{3}+\frac{4^{n+1}-4^n}{3}=
=\frac{1}{3}(n\cdot 4^{n+1}-4-4^2-4^3-...-4^n)=\frac{1}{3}\(n\cdot 4^{n+1}-\frac{4(1-4^n)}{1-4}\)=
=\frac{1}{3}\(n\cdot 4^{n+1}-\frac{4^{n+1}-4}{3}\)=\frac{(3n-1)4^{n+1}+4}{9}\)
lub
\(f(x)=\sum_{k=0}^{n}k\cdot x^k=x\sum_{k=1}^{n}k\cdot x^{k-1}=x\sum_{k=1}^{n}\(x^{k}\)'=x\(\sum_{k=1}^{n}x^{k}\)'=x\(x\cdot\frac{1-x^n}{1-x}\)'=x\(\frac{x-x^{n+1}}{1-x}\)'=\\=x\cdot\frac{(1-(n+1)x^{n})(1-x)-(x-x^{n+1})\cdot(-1)}{(1-x)^2}=x\cdot\frac{1-x-(n+1)x^{n}+(n+1)x^{n+1}+x-x^{n+1}}{(1-x)^2}=\\=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(1-x)^2}
f(4)=\sum_{k=0}^{n}k\cdot 4^k=\frac{n\cdot 4^{n+2}-(n+1)\cdot 4^{n+1}+4}{(1-4)^2}=\frac{4n\cdot 4^{n+1}-(n+1)\cdot 4^{n+1}+4}{9}=\frac{(3n-1)\cdot 4^{n+1}+4}{9}\)