Sequences and series (10)

Algebra Level 4

∑ r = 1 117 1 2 ⌊ r ⌋ + 1 = a b \large \sum_{r=1}^{117} \dfrac{1}{2 \left \lfloor \sqrt{r} \right \rfloor +1} = \frac ab

If the equation above holds true for positive coprime integers a a and b b , enter a + b + 2 a+b +2 as your answer.

Notation: ⌊ ⋅ ⌋ \left \lfloor \cdot \right \rfloor denotes the floor function .

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The answer is 78.

Rahil Sehgal by Rahil Sehgal , 20, India

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3 solutions

Tapas Mazumdar
May 9, 2017

S = ∑ r = 1 117 1 2 ⌊ r ⌋ + 1 = ∑ r = 1 117 ∫ 0 1 x 2 ⌊ r ⌋ d x = ∫ 0 1 ( ∑ r = 1 117 x 2 ⌊ r ⌋ ) d x = ∫ 0 1 ( ∑ n = 1 9 ( 2 n + 1 ) x 2 n + 18 ⋅ x 20 ) d x = ∑ n = 1 9 ∫ 0 1 ( 2 n + 1 ) x 2 n d x + ∫ 0 1 18 ⋅ x 20 d x = ∑ n = 1 9 ( 1 ) + 18 21 = 9 + 6 7 = 69 7 ⟹ a + b + 2 = 69 + 7 + 2 = 78 \begin{aligned} \displaystyle S &= \sum_{r=1}^{117} \dfrac{1}{2 \left\lfloor \sqrt{r} \right\rfloor +1} \\ &= \displaystyle \sum_{r=1}^{117} \int_0^1 x^{2 \left\lfloor \sqrt{r} \right\rfloor} \mathrm{d}x \\ &= \displaystyle \int_0^1 \left( \sum_{r=1}^{117} x^{2 \left\lfloor \sqrt{r} \right\rfloor} \right) \mathrm{d}x \\ &= \displaystyle \int_0^1 \left( \sum_{n=1}^9 (2n+1) x^{2n} + 18 \cdot x^{20} \right) \mathrm{d}x \\ &= \displaystyle \sum_{n=1}^9 \int_0^1 (2n+1) x^{2n} \mathrm{d}x + \int_0^1 18 \cdot x^{20} \mathrm{d}x \\ &= \displaystyle \sum_{n=1}^9 (1) + \dfrac{18}{21} \\ &= 9 + \dfrac 67 \\ &= \dfrac{69}{7} \end{aligned} \\ \implies a+b+2 = 69+7+2 = \boxed{78}

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How did you convert it into integral?

Sahil Silare - 4 years ago

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Simply by observing

∫ 0 1 x n d x = 1 n + 1 \int_0^1 x^n \ \mathrm{d} x = \dfrac{1}{n+1}

we can put n = 2 ⌊ r ⌋ n = 2 \left\lfloor r \right\rfloor .

Tapas Mazumdar - 4 years ago

I didnt understand your 4th step. How did the summation limit got changed to 1 to 9.

ritik agrawal - 4 years ago

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Simply observing that if r r lies between two perfect squares k 2 k^2 and ( k + 1 ) 2 (k+1)^2 then

⌊ r ⌋ = k for r ∈ [ k 2 , ( k + 1 ) 2 − 1 ] \left\lfloor \sqrt{r} \right\rfloor = k \quad \text{ for} \ r \in \left[k^2 , (k+1)^2 - 1 \right]

Now between k 2 k^2 and ( k + 1 ) 2 − 1 (k+1)^2 - 1 , there are a total of 2 k + 1 2k+1 integers. Thus between a perfect square and the preceding integer of the next perfect square we'll get 2 k + 1 2k+1 terms of 2 ⌊ r ⌋ = 2 k 2 \left\lfloor \sqrt{r} \right\rfloor = 2k .

Further observe that these intervals are

[ 1 , 3 ] ; [ 4 , 8 ] ; [ 9 , 15 ] ; [ 16 , 24 ] ; [ 25 , 35 ] ; [ 36 , 48 ] ; [ 49 , 63 ] ; [ 64 , 80 ] ; [ 81 , 99 ] [1,3] ; [4,8] ; [9,15] ; [16,24] ; [25,35] ; [36,48] ; [49,63] ; [64,80] ; [81,99]

With each respective interval giving us the value of ⌊ r ⌋ \left\lfloor \sqrt{r} \right\rfloor from 1 to 9. The last interval, i.e., from [ 100 , 117 ] [100,117] isn't exactly ending as a preceding digit of the next perfect square (which in this case would be 120), so we can just simply count the number of these which is 18 18 and here ⌊ r ⌋ = 10 ⟹ 2 ⌊ r ⌋ = 20 \left\lfloor \sqrt{r} \right\rfloor = 10 \implies 2 \left\lfloor \sqrt{r} \right\rfloor =20 .

Tapas Mazumdar - 4 years ago

S = ∑ r = 1 117 1 2 ⌊ r ⌋ + 1 = 1 3 ⏟ r = 1 2 + 1 3 + 1 3 ⏞ 2 2 − 1 2 terms + 1 5 ⏟ r = 2 2 + 1 5 + . . . + 1 5 ⏞ 3 2 − 2 2 terms + 1 7 ⏟ r = 3 2 + 1 7 + . . . + 1 7 ⏞ 4 2 − 3 2 terms + . . . + 1 21 ⏟ r = 1 0 2 + 1 21 + . . . + 1 21 = ∑ k = 1 9 ( k + 1 ) 2 − k 2 2 k + 1 + 18 21 = ∑ k = 1 9 2 k + 1 2 k + 1 + 6 7 = ∑ k = 1 9 1 + 6 7 = 9 + 6 7 = 69 7 \begin{aligned} S & = \sum_{r=1}^{117} \frac 1{2\left \lfloor \sqrt r \right \rfloor + 1} \\ & = \overbrace{\underbrace{\frac 13}_{r=1^2} + \frac 13 + \frac 13}^{2^2-1^2 \text{ terms}} + \overbrace{\underbrace{\frac 15}_{r=2^2} + \frac 15 + ... + \frac 15}^{3^2-2^2 \text{ terms}} + \overbrace{\underbrace{\frac 17}_{r=3^2} + \frac 17 + ... + \frac 17}^{4^2-3^2 \text{ terms}} + ... + \underbrace{\frac 1{21}}_{r=10^2} + \frac 1{21} + ... + \frac 1{21} \\ & = \sum_{k=1}^{9} \frac {(k+1)^2-k^2}{2k+1} + \frac {18}{21} \\ & = \sum_{k=1}^{9} \frac {2k+1}{2k+1} + \frac 67 \\ & = \sum_{k=1}^{9} 1 + \frac 67 \\ & = 9 + \frac 67 \\ & = \frac {69}7 \end{aligned}

⟹ a + b + 2 = 69 + 7 + 2 = 78 \implies a+b+2 = 69+7+2 = \boxed{78}

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Guilherme Niedu
May 9, 2017

S = ∑ r = 1 117 1 2 ⌊ r ⌋ + 1 \large \displaystyle S = \sum_{r=1}^{117} \frac{1}{2 \lfloor \sqrt{r} \rfloor + 1}

S = 1 3 + 1 3 + 1 3 + 1 5 + 1 5 + 1 5 + 1 5 + 1 5 + . . . + 1 21 \large \displaystyle S = \frac13 + \frac13 + \frac13 + \frac15 + \frac15 + \frac15 + \frac15 + \frac15 + ... + \frac{1}{21}

S = 3 ⋅ 1 3 + 5 ⋅ 1 5 + 7 ⋅ 1 7 + 9 ⋅ 1 9 + 11 ⋅ 1 11 + 13 ⋅ 1 13 + 15 ⋅ 1 15 + 17 ⋅ 1 17 + 19 ⋅ 1 19 + 18 ⋅ 1 21 \large \displaystyle S = 3\cdot \frac13 + 5\cdot \frac15 + 7\cdot \frac17 + 9\cdot \frac19 + 11\cdot \frac{1}{11} + 13\cdot \frac{1}{13} + 15\cdot \frac{1}{15} + 17\cdot \frac{1}{17} + 19\cdot \frac{1}{19} + 18\cdot \frac{1}{21}

S = 6 + 6 7 \large \displaystyle S = 6 + \frac67

S = 69 7 \color{#20A900} \boxed{ \large \displaystyle S = \frac{69}{7}}

So:

a = 69 , b = 7 , a + b + 2 = 78 \color{#3D99F6} a = 69, b = 7, \boxed{ \large \displaystyle a+b+2 = 78}

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