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Question Number 103622 by bobhans last updated on 16/Jul/20

Given a = Σ_(n=1) ^(24) (1/((√(n+1))+(√n))) then the value of a + (1/(log _a (bc)+1)) + (1/(log _b (ac)+1)) + (1/(log _c (ab)+1)) = ?

$${Given}\:{a}\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{24}} {\sum}}\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}}\:{then}\:{the}\:{value}\:{of} \\ $$$${a}\:+\:\frac{\mathrm{1}}{\mathrm{log}\:_{{a}} \left({bc}\right)+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{log}\:_{{b}} \left({ac}\right)+\mathrm{1}}\:+ \\ $$$$\frac{\mathrm{1}}{\mathrm{log}\:_{{c}} \left({ab}\right)+\mathrm{1}}\:=\:? \\ $$

Answered by OlafThorendsen last updated on 16/Jul/20

(1/((√(n+1))+(√n))) = (√(n+1))−(√n) Then a = Σ_(n=1) ^(24) ((√(n+1))−(√n)) = (√(25)) = 5 5+(1/(log_5 (bc)+1))+(1/(log_b (5c)+1))+(1/(log_c (5b)+1)) 5+(1/(((ln(bc))/(ln5))+1))+(1/(((ln(5c))/(lnb))+1))+(1/(((ln(5b))/(lnc))+1)) 5+((ln5)/(ln(bc)+ln5))+((lnb)/(ln(5c)+lnb))+((lnc)/(ln(5b)+lnc)) 5+((ln5)/(ln(5bc)))+((lnb)/(ln(5bc)))+((lnc)/(ln(5bc))) 5+((ln(5bc))/(ln(5bc))) = 5+1 = 6

$$\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}}\:=\:\sqrt{{n}+\mathrm{1}}−\sqrt{{n}} \\ $$$$\mathrm{Then}\:{a}\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{24}} {\sum}}\left(\sqrt{{n}+\mathrm{1}}−\sqrt{{n}}\right)\:=\:\sqrt{\mathrm{25}}\:=\:\mathrm{5} \\ $$$$\mathrm{5}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{5}} \left({bc}\right)+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{log}_{{b}} \left(\mathrm{5}{c}\right)+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{log}_{{c}} \left(\mathrm{5}{b}\right)+\mathrm{1}} \\ $$$$\mathrm{5}+\frac{\mathrm{1}}{\frac{\mathrm{ln}\left({bc}\right)}{\mathrm{ln5}}+\mathrm{1}}+\frac{\mathrm{1}}{\frac{\mathrm{ln}\left(\mathrm{5}{c}\right)}{\mathrm{ln}{b}}+\mathrm{1}}+\frac{\mathrm{1}}{\frac{\mathrm{ln}\left(\mathrm{5}{b}\right)}{\mathrm{ln}{c}}+\mathrm{1}} \\ $$$$\mathrm{5}+\frac{\mathrm{ln5}}{\mathrm{ln}\left({bc}\right)+\mathrm{ln5}}+\frac{\mathrm{ln}{b}}{\mathrm{ln}\left(\mathrm{5}{c}\right)+\mathrm{ln}{b}}+\frac{\mathrm{ln}{c}}{\mathrm{ln}\left(\mathrm{5}{b}\right)+\mathrm{ln}{c}} \\ $$$$\mathrm{5}+\frac{\mathrm{ln5}}{\mathrm{ln}\left(\mathrm{5}{bc}\right)}+\frac{\mathrm{ln}{b}}{\mathrm{ln}\left(\mathrm{5}{bc}\right)}+\frac{\mathrm{ln}{c}}{\mathrm{ln}\left(\mathrm{5}{bc}\right)} \\ $$$$\mathrm{5}+\frac{\mathrm{ln}\left(\mathrm{5}{bc}\right)}{\mathrm{ln}\left(\mathrm{5}{bc}\right)}\:=\:\mathrm{5}+\mathrm{1}\:=\:\mathrm{6} \\ $$$$ \\ $$

Commented by bobhans last updated on 16/Jul/20

cooll

$${cooll} \\ $$

Commented by bemath last updated on 16/Jul/20

wrong sir Σ_(n=1) ^(24) (√(n+1))−(√n) = 4

$${wrong}\:{sir}\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{24}} {\sum}}\:\sqrt{{n}+\mathrm{1}}−\sqrt{{n}}\:=\:\mathrm{4}\: \\ $$

Commented by OlafThorendsen last updated on 16/Jul/20

exact sir. you are right.

$$\mathrm{exact}\:\mathrm{sir}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}. \\ $$

Answered by bemath last updated on 16/Jul/20

a+(1/(log _a (abc))) + (1/(log _b (abc))) +(1/(log _c (abc)))= a+ log _(abc) (abc) = a+1 where a = Σ_(n=1) ^(24) (√(n+1)) −(√n) = 5−1 = 4 then ⇔ a+1 = 4+1 = 5

$${a}+\frac{\mathrm{1}}{\mathrm{log}\:_{{a}} \left({abc}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{log}\:_{{b}} \left({abc}\right)}\:+\frac{\mathrm{1}}{\mathrm{log}\:_{{c}} \left({abc}\right)}= \\ $$$${a}+\:\mathrm{log}\:_{{abc}} \left({abc}\right)\:=\:{a}+\mathrm{1}\: \\ $$$${where}\:{a}\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{24}} {\sum}}\sqrt{{n}+\mathrm{1}}\:−\sqrt{{n}}\:=\:\mathrm{5}−\mathrm{1}\:=\:\mathrm{4} \\ $$$${then}\:\Leftrightarrow\:{a}+\mathrm{1}\:=\:\mathrm{4}+\mathrm{1}\:=\:\mathrm{5}\: \\ $$

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