Exams Joint Entrance Examination Mathematics Sequences and Series Question 102 Question 102 of 519 Hard NUMERICAL
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+(\sqrt{\frac{b}{a}}+1){\log }_{e}(\frac{a}{b}) ∞ = 2 + ( a b + 1 ) log e ( b a ) , where a and b are integers with g c d ( a , b ) = 1 gcd(a,b)=1 g c d ( a , b ) = 1 , then 11 a + 18 b 11a+18b 11 a + 18 b is equal to \_\_\_\_\_\_\_\_\_\_.
Answer: 76
Solutions · 1 Post your solution Official solution
S = 1 + 3 − 2 2 3 + 5 − 2 6 18 + 9 3 − 11 2 36 3 + … ∞ = 1 + ( 1 − 2 / 3 ) 2 + ( 1 − 2 / 3 ) 2 6 + ( 1 − 2 / 3 ) 3 12 + … ∞ \begin{gathered} S=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}}+\ldots \infty \\ =1+\frac{(1-\sqrt{2}/\sqrt{3})}{2}+\frac{(1-\sqrt{2}/\sqrt{3}{)}^{2}}{6}+\frac{(1-\sqrt{2}/\sqrt{3}{)}^{3}}{12}+\ldots \infty \end{gathered} S = 1 + 2 3 3 − 2 + 18 5 − 2 6 + 36 3 9 3 − 11 2 + … ∞ = 1 + 2 ( 1 − 2 / 3 ) + 6 ( 1 − 2 / 3 ) 2 + 12 ( 1 − 2 / 3 ) 3 + … ∞
1 − 2 3 = a 1-\frac{\sqrt{2}}{\sqrt{3}}=a 1 − 3 2 = a
S = 1 + a 2 + a 2 6 + a 3 12 + … = 1 + ( 1 − 1 2 ) a + ( 1 2 − 1 3 ) a 2 + ( 1 3 − 1 4 ) a 3 + … = 1 + ( a + a 2 2 + a 3 3 … ∞ ) + 1 a ( − a 2 2 − a 3 3 − a 4 4 … ∞ ) = − ln ( 1 − a ) + 1 a ( − a − a 2 2 − a 3 3 … ∞ ) + 2 = − ln ( 1 − a ) + 1 a ln ( 1 − a ) + 2 = 2 + ( 1 a − 1 ) ln ( 1 − a ) = 2 + ( 3 3 − 2 − 1 ) ln ( 1 − 1 + 2 3 ) = 2 + 2 3 − 2 ln 2 3 = 2 + ( 6 + 2 1 ⋅ 1 2 ln 2 3 ) ∴ 2 + ( 3 2 + 1 ) ln 2 3 ∴ 11 a + 18 b = 76 \begin{gathered} S=1+\frac{a}{2}+\frac{{a}^{2}}{6}+\frac{{a}^{3}}{12}+\ldots \\ =1+(1-\frac{1}{2})a+(\frac{1}{2}-\frac{1}{3}){a}^{2}+(\frac{1}{3}-\frac{1}{4}){a}^{3}+\ldots \\ =1+(a+\frac{{a}^{2}}{2}+\frac{{a}^{3}}{3}\ldots \infty)+\frac{1}{a}(\frac{-{a}^{2}}{2}-\frac{{a}^{3}}{3}-\frac{{a}^{4}}{4}\ldots \infty) \\ =-\ln (1-a)+\frac{1}{a}(-a-\frac{{a}^{2}}{2}-\frac{{a}^{3}}{3}\ldots \infty)+2 \\ =-\ln (1-a)+\frac{1}{a}\ln (1-a)+2 \\ =2+(\frac{1}{a}-1)\ln (1-a) \\ =2+(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-1)\ln (1-1+\sqrt{\frac{2}{3}}) \\ =2+\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{2}\ln \sqrt{\frac{2}{3}} \\ =2+(\frac{\sqrt{6}+2}{1}\cdot \frac{1}{2}\ln \frac{2}{3}) \\ \therefore 2+(\sqrt{\frac{3}{2}}+1)\ln \frac{2}{3} \\ \therefore 11a+18b=76 \end{gathered} S = 1 + 2 a + 6 a 2 + 12 a 3 + … = 1 + ( 1 − 2 1 ) a + ( 2 1 − 3 1 ) a 2 + ( 3 1 − 4 1 ) a 3 + … = 1 + ( a + 2 a 2 + 3 a 3 … ∞ ) + a 1 ( 2 − a 2 − 3 a 3 − 4 a 4 … ∞ ) = − ln ( 1 − a ) + a 1 ( − a − 2 a 2 − 3 a 3 … ∞ ) + 2 = − ln ( 1 − a ) + a 1 ln ( 1 − a ) + 2 = 2 + ( a 1 − 1 ) ln ( 1 − a ) = 2 + ( 3 − 2 3 − 1 ) ln ( 1 − 1 + 3 2 ) = 2 + 3 2 − 2 ln 3 2 = 2 + ( 1 6 + 2 ⋅ 2 1 ln 3 2 ) ∴ 2 + ( 2 3 + 1 ) ln 3 2 ∴ 11 a + 18 b = 76 Answer the question to unlock