Summation

Algebra Level 5

$\sum_{i=1}^{20} \frac{ i^5 + i^3} { i^4 + i^2 + 1 } = \dfrac{1^5 + 1^3}{1^{4} + 1^{2} + 1} + \dfrac{2^5 + 2^3}{2^{4} + 2^{2} + 1} + \dfrac{3^5+3^3}{3^{4} + 3^{2} + 1} + \dfrac{4^5+4^3}{4^{4} + 4^{2} + 1} +...$

The above 20 term summation can be expressed as $\frac{A}{B}$ , for $A, B$ are coprime integers. Find $A + B$ .

The answer is 88621.

1 solution

Rohan Gupta
Dec 19, 2014

Look Closely You'll find the nth term is

$\frac{n^{5} + n^{3}}{n^{4} +n^{2} + 1}$

Splitting it as follows-

$n - \frac{n}{n^{4}+n^{2}+1}$

Summing(we will sum the two parts separately)

First we take $\frac{n}{n^{4}+n^{2}+1}$ , which can be written as -

$\frac{1}{2}*(\frac{1}{n^{2} -n +1} - \frac{1}{n^{2} +n +1})$

Which is a telescoping series and is equal to -

$\frac{1}{2}*(\frac{1}{1} - \frac{1}{n^{2} + n + 1})$

put n = 20

$= \frac{210}{421}$

Now sum n

$= \frac{n(n+1)}{2}$

put n = 20

=210

Finally subtract

$210 - \frac{210}{421}$

$= \frac{88200}{421}$

Now add 421 and 88200 = 88621

P.S. - Was a very tough one (sigh)

you have given the wrong question...in the question (in the numerator)..i is raised to the power of 2 whereas the answer is being done with n raised to the power of 3...

Naimesh Pramanik - 4 years, 6 months ago

Summation

The answer is 88621.

1 solution

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