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Compute \(\dfrac{2 + 4 + 6 + \dots + 4004}{1 + 3 + 5 + \dots + 4003}\)

 May 10, 2020
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You have two arithmetic series:   2 + 4 + 6 + ... + 4004   and   1 + 3 + 5 + ... + 4003.

Both of them can be handled in the same way.

 

t1  is the first term

tn  is the nth term (or the last term)

n  is the number of terms

d  is the common difference between successive terms

S  is the sum of the terms.

 

For the series in the numerator:  2 + 4 + 6 + ... + 4004 

t1  =  2

tn  =  4004

n  is unknown

d  =  2

S  is unknown

 

To find n:  use this formula:  tn  =  t1 + (n - 1)·d

                                        4004  =  2 + (n - 1)·2

                                        4002  =  2n - 2

                                        4004  =  2n

                                              n  =  2002                 [there are 2002 terms in this series]

 

To find S:  use this formula:  S  =  n·(t1 + tn) / 2

                                              S  =  2002·(2 + 4004) / 2 

                                              S  =  2002·(4006) / 2

                                              S  =  4,010,006

 

Now, you need to do the same steps for the denominator and find the answer when the numerator is divided by the denominator.

 May 10, 2020

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