A series sum

Algebra Level 3

The i i th term of a series is 1 i ( i + 1 ) ( i + 2 ) ( i + 3 ) \dfrac{1}{i(i+1)(i+2)(i+3)} . What is the sum of the first n n terms of the series?

1 18 − 1 3 ( n + 1 ) ( n + 2 ) ( n + 3 ) \frac{1}{18} - \frac{1}{3(n+1)(n+2)(n+3)} 1 2 n − 1 n 2 \dfrac{1}{2^n}-\dfrac{1}{n^2} 1 18 − 1 n \dfrac{1}{18}-\dfrac{1}{n} 1 18 − 1 3 ( n + 3 ) ( n + 4 ) \dfrac{1}{18}-\dfrac{1}{3(n+3)(n+4)}

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2 solutions

Chew-Seong Cheong
Aug 20, 2019

S = ∑ i = 1 n 1 i ( i + 1 ) ( i + 2 ) ( i + 3 Decompose into partial fractions = ∑ i = 1 n 1 6 ( 1 i − 3 i + 1 + 3 i + 2 − 1 i + 3 ) = 1 6 ( 1 1 − 1 n + 1 − 2 2 + 2 n + 2 + 1 3 − 1 n + 3 ) = 1 18 − 1 3 ( n + 1 ) ( n + 2 ) ( n + 3 ) \begin{aligned} S & = \sum_{i=1}^n \frac 1{i(i+1)(i+2)(i+3} & \small \color{#3D99F6} \text{Decompose into partial fractions} \\ & = \sum_{i=1}^n \frac 16 \left(\frac 1i - \frac 3{i+1} + \frac 3{i+2} - \frac 1{i+3} \right) \\ & = \frac 16 \left(\frac 11 - \frac 1{n+1} - \frac 22 + \frac 2{n+2} + \frac 13 - \frac 1{n+3}\right) \\ & = \boxed{\frac 1{18} - \frac 1{3(n+1)(n+2)(n+3)}} \end{aligned}

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@Alak Bhattacharya , you don't need so many separate \ ( \ ) \backslash( \ \backslash) , just use one. If not it will take me hours to complete my solution above and I am quite the inventor of LaTex is smarter than that. Include all valuables ( i , n i, n , constants and operators + - \times \div = + − × ÷ = + - \times \div = . Notice the different in size of + - = within and without LaTex. Use small z z not big Z for complex number and \bar z z ˉ \bar z for its conjugate. Use \sin \alpha, \cos \beta, \tan \delta, \sec \frac \pi 2, \csc \frac \pi{12}, \cot \frac {3\pi}4 sin ⁡ α , cos ⁡ β , tan ⁡ δ , sec ⁡ π 2 , csc ⁡ π 12 , cot ⁡ 3 π 4 \sin \alpha, \cos \beta, \tan \delta, \sec \frac \pi 2, \csc \frac \pi{12}, \cot \frac {3\pi}4 and small letter instead of Sin, Cos, Tan...

Chew-Seong Cheong - 1 year, 9 months ago

∑ i = 1 n 1 ∏ p = i i + 3 p ⇒ 1 18 − 1 3 ( n + 1 ) ( n + 2 ) ( n + 3 ) \sum _{i=1}^n \frac{1}{\prod _{p=i}^{i+3} p} \Rightarrow \frac{1}{18}-\frac{1}{3 (n+1) (n+2) (n+3)}

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