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Question Number 147535 by tabata last updated on 21/Jul/21

Commented by tabata last updated on 21/Jul/21

$$provethat$$

Answered by mathmax by abdo last updated on 21/Jul/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{a}}^{\mathrm{n}}\mathrm{with}\mid \mathrm{a}\mid <1\mathrm{we}\mathrm{have}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{na}}^{\mathrm{n}-1}\Rightarrow {\mathrm{af}}^{{}^{\prime }}\left(\mathrm{a}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{na}}^{\mathrm{n}}\mathrm{by}\mathrm{derivation}\mathrm{we}\mathrm{get}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)+{\mathrm{af}}^{\left(2\right)}\left(\mathrm{a}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{n}}^2{\mathrm{a}}^{\mathrm{n}-1}\Rightarrow {\mathrm{af}}^{{}^{\prime }}\left(\mathrm{a}\right)+{\mathrm{a}}^2{\mathrm{f}}^{\left(2\right)}\left(\mathrm{a}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{n}}^2{\mathrm{a}}^{\mathrm{n}}$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{{\mathrm{a}}^{\mathrm{n}+1}-1}{\mathrm{a}-1}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{{\mathrm{na}}^{\mathrm{n}+1}-(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}}+1}{{(\mathrm{a}-1)}^2}\Rightarrow$$$${\mathrm{f}}^{\left(2\right)}\left(\mathrm{a}\right)=\frac{\mathrm{n}(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}}-\mathrm{n}(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}-1}){(\mathrm{a}-1)}^2-2(\mathrm{a}-1)({\mathrm{na}}^{\mathrm{n}+1}-(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}}+1)}{{(\mathrm{a}-1)}^4}$$$$=\frac{\mathrm{n}(\mathrm{n}+1)({\mathrm{a}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}-1})(\mathrm{a}-1)-{2\mathrm{na}}^{\mathrm{n}+1}+2(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}}-2}{{(\mathrm{a}-1)}^3}$$$$=\frac{\mathrm{n}(\mathrm{n}+1)({\mathrm{a}}^{\mathrm{n}+1}-{\mathrm{a}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}}+{\mathrm{a}}^{\mathrm{n}-1})-{2\mathrm{na}}^{\mathrm{n}+1}+2(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}}-2}{{(\mathrm{a}-1)}^3}$$$$=\frac{({\mathrm{n}}^2+\mathrm{n}-2\mathrm{n}){\mathrm{a}}^{\mathrm{n}+1}+(-{2\mathrm{n}}^2-2\mathrm{n}+2\mathrm{n}+2){\mathrm{a}}^{\mathrm{n}}+\mathrm{n}(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}-1}-2}{{(\mathrm{a}-1)}^3}$$$$=\frac{({\mathrm{n}}^2-\mathrm{n}){\mathrm{a}}^{\mathrm{n}+1}-2({\mathrm{n}}^2-1){\mathrm{a}}^{\mathrm{n}}+\mathrm{n}(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}-1}-2}{{(\mathrm{a}-1)}^3}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{n}}^2{\mathrm{a}}^{\mathrm{n}}=\frac{\mathrm{a}}{{(\mathrm{a}-1)}^2}({\mathrm{na}}^{\mathrm{n}+1}-(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}}+1)$$$$-\frac{{\mathrm{a}}^2}{{(1-\mathrm{a})}^3}\{({\mathrm{n}}^2-\mathrm{n}){\mathrm{a}}^{\mathrm{n}+1}-2({\mathrm{n}}^2-1){\mathrm{a}}^{\mathrm{n}}+\mathrm{n}(\mathrm{n}+1){\mathrm{a}}^{\mathrm{n}-1}-2\}$$$$\mathrm{rest}\mathrm{to}\mathrm{simplify}\mathrm{calculus}...$$

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