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Question Number 173377 by Shrinava last updated on 10/Jul/22

If a,b,c,d>0 And (1/(a+1)) + (2/(b+1)) + (3/(c+1)) + (4/(d+1)) = 1 Then find (a/(a+1)) + ((2b)/(b+1)) + ((3c)/(c+1)) + ((4d)/(d+1))

$$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}>\mathrm{0} \\ $$ $$\mathrm{And}\:\:\frac{\mathrm{1}}{\mathrm{a}+\mathrm{1}}\:+\:\frac{\mathrm{2}}{\mathrm{b}+\mathrm{1}}\:+\:\frac{\mathrm{3}}{\mathrm{c}+\mathrm{1}}\:+\:\frac{\mathrm{4}}{\mathrm{d}+\mathrm{1}}\:=\:\mathrm{1} \\ $$ $$\mathrm{Then}\:\mathrm{find}\:\:\frac{\mathrm{a}}{\mathrm{a}+\mathrm{1}}\:+\:\frac{\mathrm{2b}}{\mathrm{b}+\mathrm{1}}\:+\:\frac{\mathrm{3c}}{\mathrm{c}+\mathrm{1}}\:+\:\frac{\mathrm{4d}}{\mathrm{d}+\mathrm{1}} \\ $$

Answered by Rasheed.Sindhi last updated on 10/Jul/22

determinant (((If a,b,c,d>0_(And (1/(a+1)) + (2/(b+1)) + (3/(c+1)) + (4/(d+1)) = 1_() ) ))) (a/(a+1)) + ((2b)/(b+1)) + ((3c)/(c+1)) + ((4d)/(d+1)) =(a/(a+1))−1 + ((2b)/(b+1))−2 + ((3c)/(c+1))−3 + ((4d)/(d+1))−4+10 =((−1)/(a+1)) + ((−2)/(b+1)) + ((−3)/(c+1)) + ((−4)/(d+1))+10 =−((1/(a+1)) + (2/(b+1)) + (3/(c+1)) + (4/(d+1)))+10 =−(1)+10=9

$$\begin{array}{|c|}{\underset{\underset{} {\mathrm{And}\:\:\frac{\mathrm{1}}{\mathrm{a}+\mathrm{1}}\:+\:\frac{\mathrm{2}}{\mathrm{b}+\mathrm{1}}\:+\:\frac{\mathrm{3}}{\mathrm{c}+\mathrm{1}}\:+\:\frac{\mathrm{4}}{\mathrm{d}+\mathrm{1}}\:=\:\mathrm{1}}} {\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}>\mathrm{0}}}\\\hline\end{array} \\ $$ $$\:\frac{\mathrm{a}}{\mathrm{a}+\mathrm{1}}\:+\:\frac{\mathrm{2b}}{\mathrm{b}+\mathrm{1}}\:+\:\frac{\mathrm{3c}}{\mathrm{c}+\mathrm{1}}\:+\:\frac{\mathrm{4d}}{\mathrm{d}+\mathrm{1}} \\ $$ $$\:=\frac{\mathrm{a}}{\mathrm{a}+\mathrm{1}}−\mathrm{1}\:+\:\frac{\mathrm{2b}}{\mathrm{b}+\mathrm{1}}−\mathrm{2}\:+\:\frac{\mathrm{3c}}{\mathrm{c}+\mathrm{1}}−\mathrm{3}\:+\:\frac{\mathrm{4d}}{\mathrm{d}+\mathrm{1}}−\mathrm{4}+\mathrm{10} \\ $$ $$\:=\frac{−\mathrm{1}}{\mathrm{a}+\mathrm{1}}\:+\:\frac{−\mathrm{2}}{\mathrm{b}+\mathrm{1}}\:+\:\frac{−\mathrm{3}}{\mathrm{c}+\mathrm{1}}\:+\:\frac{−\mathrm{4}}{\mathrm{d}+\mathrm{1}}+\mathrm{10} \\ $$ $$\:=−\left(\frac{\mathrm{1}}{\mathrm{a}+\mathrm{1}}\:+\:\frac{\mathrm{2}}{\mathrm{b}+\mathrm{1}}\:+\:\frac{\mathrm{3}}{\mathrm{c}+\mathrm{1}}\:+\:\frac{\mathrm{4}}{\mathrm{d}+\mathrm{1}}\right)+\mathrm{10} \\ $$ $$=−\left(\mathrm{1}\right)+\mathrm{10}=\mathrm{9} \\ $$

Commented bymr W last updated on 10/Jul/22

nice solution sir!

$${nice}\:{solution}\:{sir}! \\ $$

Commented byRasheed.Sindhi last updated on 10/Jul/22

Thanks a lot sir!

$$\mathcal{T}{hanks}\:{a}\:{lot}\:{sir}! \\ $$

Commented byShrinava last updated on 10/Jul/22

perfect professor thank you

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

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