Exams Joint Entrance Examination Mathematics Sequences and Series Question 55 Question 55 of 519 Medium MCQ
If 1 1 + 2 + 1 2 + 3 + … + 1 99 + 100 = m \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots +\frac{1}{\sqrt{99}+\sqrt{100}}=m 1 + 2 1 + 2 + 3 1 + … + 99 + 100 1 = m and 1 1 ⋅ 2 + 1 2 ⋅ 3 + … + 1 99 ⋅ 100 = n \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots +\frac{1}{99\cdot 100}=n 1 ⋅ 2 1 + 2 ⋅ 3 1 + … + 99 ⋅ 100 1 = n , then the point ( m , n ) (m,n) ( m , n ) lies on the line
Solutions · 1 Post your solution Official solution
1 1 + 2 + 1 2 + 3 + … + 1 99 + 100 = m 1 1 ⋅ 2 + 1 2 ⋅ 3 + … + 1 99 ⋅ 100 = n 1 1 + 2 + 1 2 + 3 + … + 1 99 + 100 = 1 1 + 2 × 2 − 1 2 − 1 + 1 3 + 2 × 3 − 2 3 − 2 + … + 1 99 + 100 × 100 − 99 100 − 99 \begin{gathered} \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots +\frac{1}{\sqrt{99}+\sqrt{100}}=m \\ \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots +\frac{1}{99\cdot 100}=n \\ \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots +\frac{1}{\sqrt{99}+\sqrt{100}} \\ =\frac{1}{\sqrt{1}+\sqrt{2}}\times \frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ \, +\ldots +\frac{1}{\sqrt{99}+\sqrt{100}}\times \frac{\sqrt{100}-\sqrt{99}}{\sqrt{100}-\sqrt{99}} \end{gathered} 1 + 2 1 + 2 + 3 1 + … + 99 + 100 1 = m 1 ⋅ 2 1 + 2 ⋅ 3 1 + … + 99 ⋅ 100 1 = n 1 + 2 1 + 2 + 3 1 + … + 99 + 100 1 = 1 + 2 1 × 2 − 1 2 − 1 + 3 + 2 1 × 3 − 2 3 − 2 + … + 99 + 100 1 × 100 − 99 100 − 99
= 2 − 1 + 3 − 2 + … + 100 − 99 = 100 − 1 = 10 − 1 ⇒ m = 9 \begin{gathered} =\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\ldots +\sqrt{100}-\sqrt{99} \\ =\sqrt{100}-\sqrt{1} \\ =10-1 \\ \Rightarrow m=9 \end{gathered} = 2 − 1 + 3 − 2 + … + 100 − 99 = 100 − 1 = 10 − 1 ⇒ m = 9
and 1 1 ⋅ 2 + 1 2 ⋅ 3 + … + 1 99 ⋅ 100 = n \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots +\frac{1}{99\cdot 100}=n 1 ⋅ 2 1 + 2 ⋅ 3 1 + … + 99 ⋅ 100 1 = n
2 − 1 1 × 2 + 3 − 2 2 × 3 + … + 100 − 99 100 × 99 = n ⇒ 1 − 1 2 + 1 2 − 1 3 + … + 1 99 − 1 100 = n ⇒ n = 1 − 1 100 ⇒ n = 99 100 ( m , n ) = ( 9 , 99 100 ) \begin{gathered} \frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\ldots +\frac{100-99}{100\times 99}=n \\ \Rightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots +\frac{1}{99}-\frac{1}{100}=n \\ \Rightarrow n=1-\frac{1}{100} \\ \Rightarrow n=\frac{99}{100} \\ (m,n)=(9,\frac{99}{100}) \end{gathered} 1 × 2 2 − 1 + 2 × 3 3 − 2 + … + 100 × 99 100 − 99 = n ⇒ 1 − 2 1 + 2 1 − 3 1 + … + 99 1 − 100 1 = n ⇒ n = 1 − 100 1 ⇒ n = 100 99 ( m , n ) = ( 9 , 100 99 )
Satisfies the line 11 x − 100 y = 0 11x-100y=0 11 x − 100 y = 0 Answer the question to unlock