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Alone

27/12/2017 at 22:13

With n is a positive number and \(a_n=\left(-1\right)^n.\dfrac{n^2+n+1}{n!}\).Find the value of S=\(a_1+a_2+........+a_{2017}\)

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Kaya Renger Coordinator 28/12/2017 at 13:57

Because n is a positive number so n has two types

+) With n is an odd number (Such as: a₁ ; a₃ ; .....)

=> \(a_n=\left(-1\right).\dfrac{n^2+n+1}{n!}\)

Because a₁, a₂ , a₃ , ...... a_n are continuous numbers

So (n - 1) is an even number

=> \(a_{n-1}=1.\dfrac{n^2+n+1}{n!}\)

With 2 continuous numbers, we can see , the total of 2 number equals to 0

\(\left(-1\right).\dfrac{n^2+n+1}{n!}+1.\dfrac{n^2+n+1}{n!}=0\)

So , with 2017 continuous number , S's value is

S = \(\left(a_1+a_2\right)+\left(a_2+a_3\right)+......+\left(a_{2015}+a_{2016}\right)+a_{2017}\)

S = \(a_{2017}=\left(-1\right)^{2017}.\dfrac{2017^2+2017+1}{2017!}=-\dfrac{2017^2+2018}{2017!}\)
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Faded 22/01/2018 at 12:41

Because n is a positive number so n has two types

+) With n is an odd number (Such as: a₁ ; a₃ ; .....)

=> $a_{n} = (- 1) . \frac{n^{2} + n + 1}{n!}$

Because a₁, a₂ , a₃ , ...... a_n are continuous numbers

So (n - 1) is an even number

=> $a_{n - 1} = 1. \frac{n^{2} + n + 1}{n!}$

With 2 continuous numbers, we can see , the total of 2 number equals to 0

$(- 1) . \frac{n^{2} + n + 1}{n!} + 1. \frac{n^{2} + n + 1}{n!} = 0$

So , with 2017 continuous number , S's value is

S = $(a_{1} + a_{2}) + (a_{2} + a_{3}) + . . . . . . + (a_{2015} + a_{2016}) + a_{2017}$

S = $a_{2017} = {(- 1)}^{2017} . \frac{2017^{2} + 2017 + 1}{2017!} = - \frac{2017^{2} + 2018}{2017!}$

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