Sum and Recursive Formula of a sequence

Algebra Level 4

Let S n S_n denote the sum of first n n terms of the sequence { a n } \{a_n\} , and a 1 = 1 , a n = 2 S n 2 2 S n − 1 a_1=1, a_n=\dfrac{2{S_n}^2}{2S_n-1} for n ≥ 2 n \geq 2 . Find S 2019 S_{2019} .

The answer can be expressed as p q \dfrac{p}{q} , where p p and q q are coprime positive integers.

Submit p + q p+q .


The answer is 4038.

Alice Smith by Alice Smith , 20, China

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2 solutions

Hassan Abdulla
Aug 18, 2019

S n = a 1 + a 2 + . . . + a n − 1 + a n S n = S n − 1 + a n ⇒ a n = S n − S n − 1 a n = 2 S n 2 2 S n − 1 = 1 2 ⋅ 4 S n 2 − 1 + 1 2 S n − 1 2 ( S n − S n − 1 ) = 4 S n 2 − 1 2 S n − 1 + 1 2 S n − 1 2 S n − 2 S n − 1 = 2 S n + 1 + 1 2 S n − 1 ⇒ S n = S n − 1 2 S n − 1 + 1 S 1 = a 1 = 1 , S 2 = S 1 2 S 1 + 1 = 1 3 , S 3 = S 2 2 S 2 + 1 = 1 5 S n = 1 2 n − 1 use induction to prove it S 2019 = 1 2 ( 2019 ) − 1 = 1 4037 \begin{aligned} S_n &= {\color{#D61F06} a_1 + a_2 + ... + a_{n-1}} +a_n \\ S_n &= {\color{#D61F06} S_{n-1} } +a_n \Rightarrow a_n = S_n - S_{n-1} \\ a_n &= \frac{2S^2_n}{2S_n - 1} = {\color{#D61F06} \frac{1}{2}} \cdot \frac{{\color{#D61F06} 4} S^2_n{\color{#3D99F6} -1 +1}}{2 S_n - 1}\\ {\color{#D61F06} 2}(S_n - S_{n-1}) &= \frac{4S^2_n-1}{2 S_n - 1} + \frac{1}{2 S_n - 1}\\ 2S_n - 2S_{n-1}&= 2S_n + 1 + \frac{1}{2 S_n - 1} \\ {\color{#D61F06} \Rightarrow} S_n &= \frac{S_{n-1}}{2S_{n-1}+1}\\ & S_1 = a_1 = 1 \, ,S_2 = \frac{S_1}{2S_1+1} = \frac{1}{3} \, ,S_3 = \frac{S_2}{2S_2+1} = \frac{1}{5}\\ & S_n=\frac{1}{2n-1} {\color{#D61F06} \text{ use induction to prove it}} \\ S_{2019}&=\frac{1}{2(2019)-1} = \frac{1}{4037} \end{aligned}

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Joël Ganesh
Aug 20, 2019

In this explanation, I refer to n n as an integer greater or equal to 2. The goal is to find two different expressions for a n a_n . One is already given, as a n = 2 S n 2 2 S n − 1 a_n = \frac{2S_n^2}{2S_n-1} . For the other one we will have to do some work. First, observe that S n = S n − 1 + a n S_n = S_{n-1} + a_n , so a n = 2 ( S n − 1 + a n ) 2 2 ( S n − 1 + a n ) − 1 , a_n = \frac{2(S_{n-1}+a_n)^2}{2(S_{n-1}+a_n)-1}, from which follows that 2 a n ( S n − 1 + a n ) − a n = 2 ( S n − 1 + a n ) 2 . 2a_n(S_{n-1}+a_n)-a_n = 2(S_{n-1}+a_n)^2. By simplifying this equation, we obtain that 2 S n − 1 2 + a n ⋅ ( 2 S n − 1 + 1 ) = 0. 2S_{n-1}^2 + a_n \cdot (2S_{n-1}+1) = 0. Solving this for a n a_n leads to the conclusion that a n = − 2 S n − 1 2 2 S n − 1 + 1 . a_n = \frac{-2S_{n-1}^2}{2S_{n-1}+1}. Now that we have found another expression for a n a_n we note that a n = 2 S n 2 2 S n − 1 = − 2 S n − 1 2 2 S n − 1 + 1 = a n . a_n = \boxed{\frac{2S_n^2}{2S_n-1} = \frac{-2S_{n-1}^2}{2S_{n-1}+1}} = a_n. Simplifying the boxed equation gives us S n 2 ( 2 S n − 1 + 1 ) = − S n − 1 2 ( 2 S n − 1 ) and so 2 S n − 1 ⋅ S n 2 + 2 S n − 1 2 ⋅ S n = S n − 1 2 − S n 2 . S_n^2(2S_{n-1}+1) = -S_{n-1}^2(2S_n-1) \text{ and so } 2S_{n-1}\cdot S_n^2 + 2S_{n-1}^2\cdot S_n = S_{n-1}^2 - S_n^2. By simplifying it even further we observe that 2 S n − 1 ⋅ S n ( S n − 1 + S n ) = ( S n − 1 + S n ) ( S n − S n − 1 ) and so 2 S n − 1 ⋅ S n = S n − S n − 1 . 2S_{n-1}\cdot S_n(S_{n-1}+S_n) = (S_{n-1}+S_n)(S_n-S_{n-1}) \text{ and so } 2S_{n-1}\cdot S_n= S_n-S_{n-1}. Now, we are able to show that S n = S n − 1 2 S n − 1 + 1 S_n = \frac{S_{n-1}}{2S_{n-1}+1} , and because S 1 = a 1 = 1 S_1 = a_1 = 1 we can prove by induction that S n = 1 2 n − 1 , S_n = \frac{1}{2n-1}, from which we conclude that S 2019 = 1 2 ⋅ 2019 − 1 , and p + q = 1 + 2 ⋅ 2019 − 1 = 4038 . S_{2019} = \frac{1}{2\cdot 2019 - 1}, \text{ and } p + q = 1 + 2 \cdot 2019 - 1 = \boxed{4038}.

1 Helpful 0 Interesting 1 Brilliant 0 Confused

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