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Algebra Level 5

∑ r = 1 n 1 r ( r + 1 ) ( r + 2 ) ( r + 3 ) = 1 a − 1 b ( n + 1 ) + 1 c ( n + 2 ) − 1 d ( n + 3 ) \sum _{r=1} ^n \frac{1}{r(r+1)(r+2)(r+3)} = \dfrac{1}{a} - \dfrac{1}{b(n+1)}+ \dfrac{1}{c(n+2)}- \dfrac{1}{d(n+3)}

If the equation above holds true for constants a a , b b , c c and d d , find a + b + c + d a+b+c+d .


The answer is 33.

Shanthanu Rai by Shanthanu Rai , 21, India

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4 solutions

Relevant wiki: Telescoping Series - Sum

A = 1 1.2.3.4 + 1 2.3.4.5 + 1 3.4.5.6 + ⋯ + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) \large \mathfrak{A} = \frac{1}{1.2.3.4} +\frac{1}{2.3.4.5} + \frac{1}{3.4.5.6} + \cdots + \frac{1}{n(n+1)(n+2)(n+3)}

3 A = 3 1.2.3.4 + 3 2.3.4.5 + 3 3.4.5.6 + ⋯ + 3 n ( n + 1 ) ( n + 2 ) ( n + 3 ) \large 3\mathfrak{A} = \frac{3}{1.2.3.4} +\frac{3}{2.3.4.5} + \frac{3}{3.4.5.6} + \cdots + \frac{3}{n(n+1)(n+2)(n+3)}

3 A = 4 − 1 1.2.3.4 + 5 − 2 2.3.4.5 + 6 − 3 3.4.5.6 + ⋯ + ( n + 3 ) − n n ( n + 1 ) ( n + 2 ) ( n + 3 ) \large 3\mathfrak{A} = \frac{4-1}{1.2.3.4} +\frac{5-2}{2.3.4.5} + \frac{6-3}{3.4.5.6} + \cdots + \frac{(n+3) - n}{n(n+1)(n+2)(n+3)}

We observe a \large \text{We observe a } Telescoping Series

3 A = 4 1.2.3. 4 − 1 1.2.3.4 + 5 2.3.4 5 − 2 2 . 3.4.5 + 6 3.4.5. 6 − 3 3 . 4.5.6 + ⋯ + ( n + 3 ) n ( n + 1 ) ( n + 2 ) ( n + 3 ) − n n ( n + 1 ) ( n + 2 ) ( n + 3 ) \large 3\mathfrak{A} = \frac{\color{#D61F06}{\cancel{4}}}{1.2.3.\color{#D61F06}{\cancel{4}}} - \frac{1}{1.2.3.4} + \frac{\color{#D61F06}{\cancel{5}}}{2.3.4\color{#D61F06}{\cancel{5}}} - \frac{\color{#D61F06}{\cancel{2}}}{\color{#D61F06}{\cancel{2}}.3.4.5} + \frac{\color{#D61F06}{\cancel{6}}}{3.4.5.\color{#D61F06}{\cancel{6}}} - \frac{\color{#D61F06}{\cancel{3}}}{\color{#D61F06}{\cancel{3}}.4.5.6} + \cdots + \frac{\cancel{(n+3)}}{n(n+1)(n+2)\cancel{(n+3)}} - \frac{\cancel{n}}{\cancel{n}(n+1)(n+2)(n+3)}

3 A = 1 1.2.3 − 1 1.2.3.4 + 1 1.2.3.4 − 1 3.4.5 + 1 3.4.5 − 1 4.5.6 + ⋯ + 1 n ( n + 1 ) ( n + 2 ) − 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) \large 3\mathfrak{A} = \frac{1}{1.2.3} - \frac{1}{1.2.3.4} + \frac{1}{1.2.3.4} -\frac{1}{3.4.5} + \frac{1}{3.4.5} - \frac{1}{4.5.6} + \cdots + \frac{1}{n(n+1)(n+2)} - \frac{1}{(n+1)(n+2)(n+3)}

3 A = 1 1.2.3 − 1 1.2.3.4 + 1 1.2.3.4 ⏟ − 1 3.4.5 + 1 3.4.5 ⏟ − 1 4.5.6 + 1 4.5.6 ⏟ ⋯ − 1 n ( n + 1 ) ( n + 2 ) + 1 n ( n + 1 ) ( n + 2 ) ⏟ − 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) \large 3\mathfrak{A} = \frac{1}{1.2.3} - \underbrace{\cancel{\frac{1}{1.2.3.4}} + \cancel{\frac{1}{1.2.3.4}}} -\underbrace{\cancel{\frac{1}{3.4.5}} + \cancel{\frac{1}{3.4.5}}} - \underbrace{\cancel{\frac{1}{4.5.6}} + \cancel{\frac{1}{4.5.6}}} \cdots - \underbrace{\cancel{\frac{1}{n(n+1)(n+2)}} + \cancel{\frac{1}{n(n+1)(n+2)}}} - \frac{1}{(n+1)(n+2)(n+3)}

All the terms as braced cancel out and yield the following : \text{All the terms as braced cancel out and yield the following : }

3 A = 1 1.2.3 − 1 2 2 ( n + 1 ) ( n + 2 ) ( n + 3 ) \large 3\mathfrak{A} = \frac{1}{1.2.3} - \color{#3D99F6}{\frac{1}{2}}\frac{\color{#3D99F6}2}{(n+1)(n+2)(n+3)}

3 A = 1 1.2.3 − 1 2 ( n + 3 ) − ( n + 1 ) ( n + 1 ) ( n + 2 ) ( n + 3 ) \large 3\mathfrak{A} = \frac{1}{1.2.3} - \frac{1}{2}\frac{(n+3)-(n+1)}{(n+1)(n+2)(n+3)}

3 A = 1 1.2.3 − 1 2 ( 1 ( n + 1 ) ( n + 2 ) − 1 ( n + 3 ) ( n + 2 ) \large 3\mathfrak{A} = \frac{1}{1.2.3} - \frac{1}{2}(\frac{1}{(n+1)(n+2)} - \frac{1}{(n+3)(n+2)}

3 A = 1 1.2.3 − 1 2 ( ( n + 2 ) − ( n + 1 ) ( n + 1 ) ( n + 2 ) − ( n + 3 ) − ( n + 2 ) ( n + 3 ) ( n + 2 ) ) \large 3\mathfrak{A} = \frac{1}{1.2.3} - \frac{1}{2}(\frac{\color{#3D99F6}{(n+2)-(n+1)}}{(n+1)(n+2)} - \frac{\color{#3D99F6}{(n+3)-(n+2)}}{(n+3)(n+2)})

3 A = 1 1.2.3 − 1 2 ( 1 ( n + 1 ) − 1 ( n + 2 ) − 1 ( n + 2 ) + 1 ( n + 3 ) \large 3\mathfrak{A} = \frac{1}{1.2.3} - \frac{1}{2}(\frac{1}{(n+1)} - \frac{1}{(n+2)} - \frac{1}{(n+2)} + \frac{1}{(n+3)}

3 A = 1 1.2.3 − ( 1 2 ( n + 1 ) − 1 2 ( n + 2 ) − 1 2 ( n + 2 ) + 1 2 ( n + 3 ) ) \large 3\mathfrak{A} = \frac{1}{1.2.3} - (\frac{1}{2(n+1)} - \frac{1}{2(n+2)} - \frac{1}{2(n+2)} + \frac{1}{2(n+3)})

3 A = 1 1.2.3 − 1 2 ( n + 1 ) + 1 ( n + 2 ) − 1 2 ( n + 3 ) ) \large 3\mathfrak{A} = \frac{1}{1.2.3} - \frac{1}{2(n+1)} + \frac{1}{(n+2)} - \frac{1}{2(n+3)})

A = 1 18 − 1 6 ( n + 1 ) + 1 3 ( n + 2 ) − 1 6 ( n + 3 ) \huge \boxed{\mathfrak{A}} = \frac{1}{18} - \frac{1}{6(n+1)} + \frac{1}{3(n+2)} - \frac{1}{6(n+3)}

⟹ a = 18 , b = 6 , c = 3 , d = 6 ⟹ a + b + c + d − 1 = 32 \implies \boxed {a=18,b=6,c=3,d=6}\implies \boxed{a+b+c+d-1=32}

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Moderator note:

In the latter half, you can directly apply partial fractions decomposition to 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) \frac{ 1}{ (n+1)(n+2)(n+3) } .

Very elaborate solution. I think there in an error in the second line when you multiply 3 to the whole equation.

Shanthanu Rai - 5 years, 2 months ago

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Yeah the typo is fixed now :)

Aditya Narayan Sharma - 5 years, 2 months ago

Oh yes ! I would keep it in mind .

Aditya Narayan Sharma - 5 years, 2 months ago

S = ∑ r = 1 n 1 r ( r + 1 ) ( r + 2 ) ( r + 3 ) By partial fraction decomposition = ∑ r = 1 n ( 1 6 r − 1 2 ( r + 1 ) + 1 2 ( r + 2 ) − 1 6 ( r + 3 ) ) = 1 6 ∑ r = 1 n ( 1 r − 1 r + 3 ) − 1 2 ∑ r = 1 n ( 1 r + 1 − 1 r + 2 ) = 1 6 ( ∑ r = 1 n 1 r − ∑ r = 4 n + 3 1 r ) − 1 2 ( ∑ r = 2 n + 1 1 r − ∑ r = 3 n + 2 1 r ) = 1 6 ( 1 + 1 2 + 1 3 − 1 n + 1 − 1 n + 2 − 1 n + 3 ) − 1 2 ( 1 2 − 1 n + 2 ) = 1 18 − 1 6 ( n + 1 ) + 1 3 ( n + 2 ) − 1 6 ( n + 3 ) \begin{aligned} S & = \sum_{r=1}^n \frac 1{r(r+1)(r+2)(r+3)} & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \sum_{r=1}^n \left(\frac 1{6r} - \frac 1{2(r+1)} + \frac 1{2(r+2)} - \frac 1{6(r+3)} \right) \\ & = \frac 16 \sum_{r=1}^n \left(\frac 1r - \frac 1{r+3} \right) - \frac 12 \sum_{r=1}^n \left( \frac 1{r+1} - \frac 1{r+2} \right) \\ & = \frac 16 \left(\sum_{r=1}^n \frac 1r - \sum_{r=4}^{n+3} \frac 1r \right) - \frac 12 \left(\sum_{r=2}^{n+1} \frac 1r - \sum_{r=3}^{n+2} \frac 1r \right) \\ & = \frac 16 \left(1+\frac 12 + \frac 13 - \frac 1{n+1} - \frac 1{n+2} - \frac 1{n+3}\right) - \frac 12 \left(\frac 12 - \frac 1{n+2}\right) \\ & = \frac 1{18} - \frac 1{6(n+1)} + \frac 1{3(n+2)} - \frac 1{6(n+3)} \end{aligned}

Therefore, a + b + c + d = 18 + 6 + 3 + 6 = 33 a+b+c+d = 18+6+3+6 = \boxed{33} .

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Hassan Abdulla
May 4, 2018

let I = ∑ r = 1 n 1 r ( r + 1 ) ( r + 2 ) ( r + 3 ) = 1 6 ∑ r = 1 n 1 r − 3 r + 1 + 3 r + 2 − 1 r + 3 I = 1 6 ∑ r = 1 n ∫ 0 1 ( x r − 1 − 3 x r + 3 x r + 1 − x r + 2 ) d x = 1 6 ∑ r = 0 n − 1 ∫ 0 1 ( x r − 3 x r + 1 + 3 x r + 2 − x r + 3 ) d x I = 1 6 ∑ r = 0 n − 1 ∫ 0 1 x r ( 1 − 3 x 1 + 3 x 2 − x 3 ) d x = − 1 6 ∑ r = 0 n − 1 ∫ 0 1 x r ( x − 1 ) 3 d x I = − 1 6 ∫ 0 1 [ ( x − 1 ) 3 ∑ r = 0 n − 1 x r ] d x = − 1 6 ∫ 0 1 [ ( x − 1 ) 3 x n − 1 x − 1 ] d x = − 1 6 ∫ 0 1 [ ( x − 1 ) 2 ( x n − 1 ) ] d x I = − 1 6 ∫ 0 1 [ x n + 2 − 2 x n + 1 + x n − ( x − 1 ) 2 ] d x = − 1 6 [ 1 n + 3 − 2 n + 2 + 1 n + 1 − 1 3 ] I = 1 18 − 1 6 ( n + 1 ) + 1 3 ( n + 2 ) − 1 6 ( n + 3 ) \large {\text{let } I=\sum _{ r=1 }^{ n }{ \frac { 1 }{ r\left( r+1 \right) \left( r+2 \right) \left( r+3 \right) } } =\frac { 1 }{ 6 } \sum _{ r=1 }^{ n }{ \frac { 1 }{ r } -\frac { 3 }{ r+1 } +\frac { 3 }{ r+2 } -\frac { 1 }{ r+3 } } \\ \large I=\frac { 1 }{ 6 } \sum _{ r=1 }^{ n }{ \int _{ 0 }^{ 1 }{ \left( { x }^{ r-1 }-3{ x }^{ r }+3{ x }^{ r+1 }-{ x }^{ r+2 } \right) dx } } =\frac { 1 }{ 6 } \sum _{ r=0 }^{ n-1 }{ \int _{ 0 }^{ 1 }{ \left( { x }^{ r }-3{ x }^{ r+1 }+3{ x }^{ r+2 }-{ x }^{ r+3 } \right) dx } } \\ I=\frac { 1 }{ 6 } \sum _{ r=0 }^{ n-1 }{ \int _{ 0 }^{ 1 }{ { x }^{ r }\left( 1-3{ x }^{ 1 }+3{ x }^{ 2 }-{ x }^{ 3 } \right) dx } } =\frac { -1 }{ 6 } \sum _{ r=0 }^{ n-1 }{ \int _{ 0 }^{ 1 }{ { x }^{ r }{ \left( x-1 \right) }^{ 3 }dx } } \\ I=\frac { -1 }{ 6 } \int _{ 0 }^{ 1 }{ \left[ { \left( x-1 \right) }^{ 3 }\sum _{ r=0 }^{ n-1 }{ { x }^{ r } } \right] dx } =\frac { -1 }{ 6 } \int _{ 0 }^{ 1 }{ \left[ { \left( x-1 \right) }^{ 3 }\frac { { x }^{ n }-1 }{ x-1 } \right] dx } =\frac { -1 }{ 6 } \int _{ 0 }^{ 1 }{ \left[ { \left( x-1 \right) }^{ 2 }\left( { x }^{ n }-1 \right) \right] dx } \\ I=\frac { -1 }{ 6 } \int _{ 0 }^{ 1 }{ \left[ { { x }^{ n+2 }-2{ x }^{ n+1 }+{ x }^{ n }-\left( x-1 \right) }^{ 2 } \right] dx } =\frac { -1 }{ 6 } \left[ \frac { 1 }{ n+3 } -\frac { 2 }{ n+2 } +\frac { 1 }{ n+1 } -\frac { 1 }{ 3 } \right] \\ I=\frac { 1 }{ 18 } -\frac { 1 }{ 6(n+1) } +\frac { 1 }{ 3(n+2) } -\frac { 1 }{ 6(n+3)}}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice approach!! +1

Shanthanu Rai - 3 years, 1 month ago

Same approach as mine...exploiting infinite geometric series + integration!

tom engelsman - 3 years, 1 month ago

well now that was pretty out of the box

Divyanshu Vadehra - 3 years ago
Shefali Sharma
Nov 4, 2018

T r T_{r} = 1 r ( r + 1 ) ( r + 2 ) ( r + 3 ) \frac{1}{r(r+1)(r+2)(r+3)}
T r + 1 T_{r+1} = \frac{1}{(r+1)(r+2)(r+3)(r+4)} \(T_{r} / T r + 1 T_{r+1} = r + 4 r \frac{r+4}{r}
r T r T_{r} = (r+4) T r + 1 T_{r+1}
r T r T_{r} = (r+1) T r + 1 T_{r+1} + 3 T r + 1 T_{r+1}
putting r = 1,2,3,.........,n-1 and adding them
T 1 T_{1} = 2 T 2 T_{2} + 3 T 2 T_{2}
2 T 2 T_{2} = 3 T 3 T_{3} +3 T 3 T_{3}
3 T 3 T_{3} = 4 T 4 T_{4} + 3 T 4 T_{4}
. .
. .
. .
(n-1) T n − 1 T_{n-1} = n T n T_{n} + 3 T n T_{n}





T 1 T_{1} = 3( T 2 T_{2} + T 3 T_{3} + ..............+ T n T_{n} ) + n T n T_{n}
4 T 1 T_{1} = 3( T 1 T_{1} + T 2 T_{2} + T 3 T_{3} +............. T n T_{n} ) + n T n T_{n} 4T1 - n T n T_{n} = 3 S n S_{n}
S n S_{n} =(4 T 1 T_{1} - n T n T_{n} )/3
S n S_{n} = 1/18 - 1/3(n+1)(n+2)(n+3)
S n S_{n} = 1/18 - 1/6(n+1) + 1/3(n+2) - 1/6(n+2)
Thus, a+b+c+d = 33

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