JEE Main 2024 5 April Evening Shift
Exam JEE Main
Year 2024
Date 5 April
Shift Evening Shift
Subject Mathematics
Chapter Sequences and Series JEE Main 2024 · 5 April · Evening Shift · Mathematics · Sequences and Series
If 1+ 3 - 2 2 3 + 5-2 6 18 + 9 3 -11 2 36 3 + 49-20 6 180 + upto...
If
1 + 3 − 2 2 3 + 5 − 2 6 18 + 9 3 − 11 2 36 3 + 49 − 20 6 180 + … 1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\frac{49-20 \sqrt{6}}{180}+\ldots 1 + 2 3 3 − 2 + 18 5 − 2 6 + 36 3 9 3 − 11 2 + 180 49 − 20 6 + … upto
∞ = 2 + ( b a + 1 ) log e ( a b ) \infty=2+\left(\sqrt{\frac{b}{a}}+1\right) \log _e\left(\frac{a}{b}\right) ∞ = 2 + ( a b + 1 ) log e ( b a ) , where
a \mathrm{a} a and
b \mathrm{b} b are integers with
gcd ( a , b ) = 1 \operatorname{gcd}(\mathrm{a}, \mathrm{b})=1 gcd ( a , b ) = 1 , then
11 a + 18 b 11 \mathrm{a}+18 \mathrm{~b} 11 a + 18 b is equal to ______
Step-by-step solution
S = 1 + x 2 3 + x 2 18 + x 3 36 3 + x 4 180 + … ∞ Put x 3 = t , where x = 3 − 2 S = 1 + t 2 + t 2 6 + t 3 12 + t 4 20 + … S = 1 + t ( 1 − 1 2 ) + t 2 ( 1 2 − 1 3 ) + t 3 ( 1 3 − 1 4 ) + t 4 ( 1 4 − 1 5 ) S = ( 1 + t + t 2 2 + t 3 3 + t 3 4 + … ) − ( t 2 + t 2 3 + t 3 4 + t 4 5 + … ) S = ( t + t 2 2 + … ) − 1 t ( t + t 2 2 + t 3 3 + … ) + 2 S = 2 + ( 1 − 1 t ) ( − log ( 1 − t ) ) = ( 1 t − 1 ) log ( 1 − t ) + 2 S = 2 + ( 3 3 − 2 − 1 ) log ( 1 − 3 − 2 3 ) S = 2 + ( 2 3 − 2 ) log e 2 3 S = 2 + ( 6 + 2 ) 2 log e 2 3 = 2 + ( 3 2 + 1 ) log e 2 3 a = 2 , b = 3 11 a + 18 b = 11 × 2 + 18 × 3 = 76 \begin{aligned} & S=1+\frac{x}{2 \sqrt{3}}+\frac{x^2}{18}+\frac{x^3}{36 \sqrt{3}}+\frac{x^4}{180}+\ldots \infty \\ & \text{ Put } \frac{x}{\sqrt{3}}=t, \text{ where } x=\sqrt{3}-\sqrt{2} \\ & S=1+\frac{t}{2}+\frac{t^2}{6}+\frac{t^3}{12}+\frac{t^4}{20}+\ldots \\ & S=1+t\left(1-\frac{1}{2}\right)+t^2\left(\frac{1}{2}-\frac{1}{3}\right)+t^3\left(\frac{1}{3}-\frac{1}{4}\right)+t^4\left(\frac{1}{4}-\frac{1}{5}\right) \\ & S=\left(1+t+\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^3}{4}+\ldots\right)-\left(\frac{t}{2}+\frac{t^2}{3}+\frac{t^3}{4}+\frac{t^4}{5}+\ldots\right) \\ & S=\left(t+\frac{t^2}{2}+\ldots\right)-\frac{1}{t}\left(t+\frac{t^2}{2}+\frac{t^3}{3}+\ldots\right)+2 \\ & S=2+\left(1-\frac{1}{t}\right)(-\log (1-t))=\left(\frac{1}{t}-1\right) \log (1-t)+2 \\ & S=2+\left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-1\right) \log \left(1-\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right) \\ & S=2+\left(\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right) \log e \frac{\sqrt{2}}{\sqrt{3}} \\ & S=2+\frac{(\sqrt{6}+2)}{2} \log e \frac{2}{3}=2+\left(\sqrt{\frac{3}{2}}+1\right) \log e \frac{2}{3} \\ & a=2, b=3 \\ & 11 a+18 b=11 \times 2+18 \times 3=76\end{aligned} S = 1 + 2 3 x + 18 x 2 + 36 3 x 3 + 180 x 4 + … ∞ Put 3 x = t , where x = 3 − 2 S = 1 + 2 t + 6 t 2 + 12 t 3 + 20 t 4 + … S = 1 + t ( 1 − 2 1 ) + t 2 ( 2 1 − 3 1 ) + t 3 ( 3 1 − 4 1 ) + t 4 ( 4 1 − 5 1 ) S = ( 1 + t + 2 t 2 + 3 t 3 + 4 t 3 + … ) − ( 2 t + 3 t 2 + 4 t 3 + 5 t 4 + … ) S = ( t + 2 t 2 + … ) − t 1 ( t + 2 t 2 + 3 t 3 + … ) + 2 S = 2 + ( 1 − t 1 ) ( − log ( 1 − t )) = ( t 1 − 1 ) log ( 1 − t ) + 2 S = 2 + ( 3 − 2 3 − 1 ) log ( 1 − 3 3 − 2 ) S = 2 + ( 3 − 2 2 ) log e 3 2 S = 2 + 2 ( 6 + 2 ) log e 3 2 = 2 + ( 2 3 + 1 ) log e 3 2 a = 2 , b = 3 11 a + 18 b = 11 × 2 + 18 × 3 = 76
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