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Question Number 96318 by  M±th+et+s last updated on 31/May/20

if  (1+x)(1+x^2 ).....(1+x^(128) )=Σ_(r=0) ^n x^r   then find n

$${if} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right).....\left(\mathrm{1}+{x}^{\mathrm{128}} \right)=\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{r}} \\ $$$${then}\:{find}\:{n} \\ $$

Commented by mr W last updated on 31/May/20

n=1+2+3+...+128=((128×129)/2)=8256

$${n}=\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{128}=\frac{\mathrm{128}×\mathrm{129}}{\mathrm{2}}=\mathrm{8256} \\ $$

Commented by  M±th+et+s last updated on 31/May/20

thanks sir

$${thanks}\:{sir} \\ $$

Commented by mr W last updated on 31/May/20

but  (1+x)(1+x^2 ).....(1+x^(128) )=Σ_(r=0) ^n a_r x^r ≠Σ_(r=0) ^n x^r

$${but} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right).....\left(\mathrm{1}+{x}^{\mathrm{128}} \right)=\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{r}} {x}^{{r}} \neq\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{r}} \\ $$

Commented by  M±th+et+s last updated on 31/May/20

yes sorry its (1+x)(1+x^2 )(1+x^4 )....(1+x^(128) )

$${yes}\:{sorry}\:{its}\:\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)....\left(\mathrm{1}+{x}^{\mathrm{128}} \right) \\ $$

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