Question Number 219761 by Spillover last updated on 01/May/25
Answered by Spillover last updated on 01/May/25
Answered by mr W last updated on 01/May/25
$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{{n}} +…=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}}\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{nx}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{nx}^{{n}} =\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} =\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$${put}\:{x}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{\mathrm{7}^{{n}−\mathrm{1}} }=\mathrm{1}+\frac{\mathrm{4}}{\mathrm{7}}+\frac{\mathrm{9}}{\mathrm{7}^{\mathrm{2}} }+\frac{\mathrm{16}}{\mathrm{7}^{\mathrm{3}} }+\frac{\mathrm{25}}{\mathrm{7}^{\mathrm{4}} }+… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{7}}\right)^{\mathrm{3}} }=\frac{\mathrm{49}}{\mathrm{27}} \\ $$
Commented by nothing48 last updated on 01/May/25
$${isn}'{t}\:{it} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{nx}^{{n}} \:{instead}\:{of}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{nx}^{{n}} \\ $$
Commented by mr W last updated on 01/May/25
$${yes},\:{but}\:{they}\:{are}\:{the}\:{same}\:{here}. \\ $$
Commented by Spillover last updated on 01/May/25
$${thank}\:{you} \\ $$