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Question Number 83931 by john santu last updated on 08/Mar/20

(1/(((√1)+(√2))((1)^(1/(4 )) +(2)^(1/(4 )) ))) + (1/(((√2)+(√3))((2)^(1/( 4)) +(3)^(1/(4 )) ))) + (1/(((√3)+(√4))((3)^(1/(4 )) +(4)^(1/(4 )) ))) + ... + (1/(((√(255))+(√(256)))(((255))^(1/(4 )) +((256))^(1/(4 )) ))) = ...

$$\frac{\mathrm{1}}{\left(\sqrt{\mathrm{1}}+\sqrt{\mathrm{2}}\right)\left(\sqrt[{\mathrm{4}\:}]{\mathrm{1}}+\sqrt[{\mathrm{4}\:}]{\mathrm{2}}\right)}\:+\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\left(\sqrt[{\:\mathrm{4}}]{\mathrm{2}}+\sqrt[{\mathrm{4}\:}]{\mathrm{3}}\right)}\:+ \\ $$$$\frac{\mathrm{1}}{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}\right)\left(\sqrt[{\mathrm{4}\:}]{\mathrm{3}}+\sqrt[{\mathrm{4}\:}]{\mathrm{4}}\right)}\:+\:...\:+\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{255}}+\sqrt{\mathrm{256}}\right)\left(\sqrt[{\mathrm{4}\:}]{\mathrm{255}}+\sqrt[{\mathrm{4}\:}]{\mathrm{256}}\right)} \\ $$$$=\:...\: \\ $$

Commented by john santu last updated on 08/Mar/20

yes. the answer 3

$$\mathrm{yes}.\:\mathrm{the}\:\mathrm{answer}\:\mathrm{3} \\ $$

Answered by TANMAY PANACEA last updated on 08/Mar/20

T_r =(1/(((√r) +(√(r+1)) )((r)^(1/4) +((r+1)))^(1/4) )) S=Σ_(r=1) ^(255) T_r T_r =(((r+1)−r)/D_r )=(((√(r+1)) −(√r))/((((r+1))^(1/4) +(r)^(1/4) )))=(((((r+1))^(1/4) +(r)^(1/4) )×(((r+1))^(1/4) −(r)^(1/4) ))/((((r+1))^(1/4) +(r)^(1/4) ))) T_r =((r+1))^(1/4) −(r)^(1/4) so Σ_(r=1) ^(255) T_r =0 pls check←it is my error so pls ignore... editing... T_r =((r+1))^(1/4) −(r)^(1/4) T_1 =(2)^(1/4) −(1)^(1/4) T_2 =(3)^(1/4) −(2)^(1/4) ... ... T_(255) =((256))^(1/4) −((255))^(1/4) S=((256))^(1/4) −(1)^(1/4) =4−1=3

$${T}_{{r}} =\frac{\mathrm{1}}{\left(\sqrt{{r}}\:+\sqrt{{r}+\mathrm{1}}\:\right)\left(\sqrt[{\mathrm{4}}]{{r}}\:+\sqrt[{\mathrm{4}}]{\left.{r}+\mathrm{1}\right)}\right.} \\ $$$${S}=\underset{{r}=\mathrm{1}} {\overset{\mathrm{255}} {\sum}}{T}_{{r}} \\ $$$${T}_{{r}} =\frac{\left({r}+\mathrm{1}\right)−{r}}{{D}_{{r}} }=\frac{\sqrt{{r}+\mathrm{1}}\:−\sqrt{{r}}}{\left(\sqrt[{\mathrm{4}}]{{r}+\mathrm{1}}\:+\sqrt[{\mathrm{4}}]{{r}}\:\right)}=\frac{\left(\sqrt[{\mathrm{4}}]{{r}+\mathrm{1}}\:+\sqrt[{\mathrm{4}}]{{r}}\:\right)×\left(\sqrt[{\mathrm{4}}]{{r}+\mathrm{1}}\:−\sqrt[{\mathrm{4}}]{{r}}\:\right)}{\left(\sqrt[{\mathrm{4}}]{{r}+\mathrm{1}}\:+\sqrt[{\mathrm{4}}]{{r}}\:\right)} \\ $$$${T}_{{r}} =\sqrt[{\mathrm{4}}]{{r}+\mathrm{1}}\:−\sqrt[{\mathrm{4}}]{{r}}\: \\ $$$${so}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{255}} {\sum}}{T}_{{r}} \:=\mathrm{0}\:\:\:\:\boldsymbol{{pls}}\:\boldsymbol{{check}}\leftarrow{it}\:{is}\:{my}\:{error} \\ $$$${so}\:{pls}\:{ignore}... \\ $$$${editing}... \\ $$$${T}_{{r}} =\sqrt[{\mathrm{4}}]{{r}+\mathrm{1}}\:−\sqrt[{\mathrm{4}}]{{r}}\: \\ $$$${T}_{\mathrm{1}} =\sqrt[{\mathrm{4}}]{\mathrm{2}}\:−\sqrt[{\mathrm{4}}]{\mathrm{1}}\: \\ $$$${T}_{\mathrm{2}} =\sqrt[{\mathrm{4}}]{\mathrm{3}}\:−\sqrt[{\mathrm{4}}]{\mathrm{2}}\: \\ $$$$... \\ $$$$... \\ $$$${T}_{\mathrm{255}} =\sqrt[{\mathrm{4}}]{\mathrm{256}}\:−\sqrt[{\mathrm{4}}]{\mathrm{255}}\: \\ $$$${S}=\sqrt[{\mathrm{4}}]{\mathrm{256}}\:−\sqrt[{\mathrm{4}}]{\mathrm{1}}\:=\mathrm{4}−\mathrm{1}=\mathrm{3} \\ $$

Commented by JDamian last updated on 08/Mar/20

if your expression for T_(r ) is correct, then S=^4 (√(255+1))−^4 (√1)=4−1=3

$${if}\:{your}\:{expression}\:{for}\:{T}_{{r}\:} \:{is}\:{correct},\:{then} \\ $$$${S}=^{\mathrm{4}} \sqrt{\mathrm{255}+\mathrm{1}}−^{\mathrm{4}} \sqrt{\mathrm{1}}=\mathrm{4}−\mathrm{1}=\mathrm{3} \\ $$

Commented by jagoll last updated on 08/Mar/20

why Σ_(r = 1) ^(255) T_r = Σ_(r = 1) ^(255) (((r+1))^(1/(4 )) − (r)^(1/(4 )) ) = 0 ? it′ s typo sir?

$$\mathrm{why}\:\underset{\mathrm{r}\:=\:\mathrm{1}} {\overset{\mathrm{255}} {\sum}}\mathrm{T}_{\mathrm{r}} \:=\:\underset{\mathrm{r}\:=\:\mathrm{1}} {\overset{\mathrm{255}} {\sum}}\left(\sqrt[{\mathrm{4}\:\:}]{\mathrm{r}+\mathrm{1}}\:−\:\sqrt[{\mathrm{4}\:\:}]{\mathrm{r}}\:\right)\:=\:\mathrm{0}\:? \\ $$$$\mathrm{it}'\:\mathrm{s}\:\mathrm{typo}\:\mathrm{sir}? \\ $$

Commented by TANMAY PANACEA last updated on 08/Mar/20

yes hurry

$${yes}\:\:{hurry} \\ $$

Commented by john santu last updated on 08/Mar/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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