Progressions

Calculus Level 3

S n = ∑ r = 1 n 2 r + 1 r 4 + 2 r 3 + r 2 S_{n}=\sum_{r=1}^n\frac{2r+1}{r^{4}+2r^{3}+r^{2}}

For S n S_n as defined above, S 20 = m m + 1 S_{20} = \dfrac{m}{m+1} . Find m + 1 \sqrt{m+1} .


The answer is 21.

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Shivam Jadhav by Shivam Jadhav , 22, India

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3 solutions

Omkar Kulkarni
Jun 2, 2015

Here's a Latex solution, for those who find the other one incomprehensible.

2 r + 1 r 4 + 2 r 3 + r 2 = 2 r + 1 r 2 ( r 2 + 2 r + 1 ) = 2 r + 1 r 2 ( r + 1 ) 2 = r 2 + 2 r + 1 − r 2 r 2 ( r + 1 ) 2 = ( r + 1 ) 2 − r 2 r 2 ( r + 1 ) 2 = ( r + 1 ) 2 r 2 ( r + 1 ) 2 − r 2 r 2 ( r + 1 ) 2 = 1 r 2 − 1 ( r + 1 ) 2 \begin{aligned} \frac{2r+1}{r^{4}+2r^{3}+r^{2}} &=& \frac{2r+1}{r^{2}(r^{2}+2r+1)} \\ &=&\frac{2r+1}{r^{2}(r+1)^{2}} \\ &=& \frac{r^{2}+2r+1-r^{2}}{r^{2}(r+1)^{2}} \\ &=& \frac{(r+1)^{2}-r^{2}}{r^{2}(r+1)^{2}} \\ &=& \frac{(r+1)^{2}}{r^{2}(r+1)^{2}}-\frac{r^{2}}{r^{2}(r+1)^{2}} \\ &=& \frac{1}{r^{2}}-\frac{1}{(r+1)^{2}} \end{aligned}

∴ ∑ r = 1 n 2 r + 1 r 4 + 2 r 3 + r 2 = ∑ r = 1 n 1 r 2 − 1 ( r + 1 ) 2 = 1 1 2 − 1 2 2 + 1 2 2 − 1 3 2 + 1 3 2 + 1 4 2 + ⋯ + 1 n 2 − 1 ( n + 1 ) 2 = 1 − 1 ( n + 1 ) 2 = n ( n + 2 ) ( n + 1 ) 2 \begin{aligned} \therefore \displaystyle \sum_{r=1}^{n} \frac{2r+1}{r^{4}+2r^{3}+r^{2}} &=& \displaystyle \sum_{r=1}^{n} \frac{1}{r^{2}}-\frac{1}{(r+1)^{2}} \\ &=& \frac{1}{1^{2}}-\frac{1}{2^{2}}+\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\dots+\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}} \\ &=& 1-\frac{1}{(n+1)^{2}} = \frac{n(n+2)}{(n+1)^{2}} \end{aligned}

∴ S 20 = ( 20 ) ( 22 ) 21 × 21 = 440 441 \therefore S_{20}=\frac{(20)(22)}{21\times21}=\boxed{\frac{440}{441}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

S n = ∑ r = 1 n 2 r + 1 r 4 + 2 r 3 + r 2 = ∑ r = 1 n 2 r + 1 r 2 ( r 2 + 2 r + 1 ) = ∑ r = 1 n 2 r + 1 r 2 ( r + 1 ) 2 = ∑ r = 1 n ( 1 r 2 − 1 ( r + 1 ) 2 ) = 1 − 1 ( n + 1 ) 2 ⟹ S 20 = 1 − 1 2 1 2 = 440 441 \begin{aligned} S_n & = \sum_{r=1}^n \frac {2r+1}{r^4+2r^3+r^2} \\ & = \sum_{r=1}^n \frac {2r+1}{r^2(r^2+2r+1)} \\ & = \sum_{r=1}^n \frac {2r+1}{r^2(r+1)^2} \\ & = \sum_{r=1}^n \left(\frac 1{r^2} - \frac 1{(r+1)^2} \right) \\ & = 1 - \frac 1{(n+1)^2} \\ \implies S_{20} & = 1 - \frac 1{21^2} = \frac {440}{441} \end{aligned}

⟹ m + 1 = 441 = 21 \implies \sqrt{m+1} = \sqrt{441} = \boxed{21}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Siddharth Iyer
Apr 21, 2015

We can factor the denominator as (r^2)((r+1)^2). Write numberator as (r+1)^2-r^2 to obtain ((r+1)^2-r^2)/((r^2)((r+1)^2) = (1/(r^2))-(1/(r+1)^2). Observe that 1/((1)^2) -1/((2)^2)+(1/((2)^2)+(1/(3)^2)...(1/(20)^2)-(1/(21^2)) = 1 - 1/(21)^2 =399/400. root(400)=20

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You made a slight mistake in the second last step. 1 − 1 2 1 2 1-\frac{1}{21^{2}} should be equal to 440 441 \frac{440}{441} , not 399 400 \frac{399}{400} .

Omkar Kulkarni - 6 years ago

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