Average of a series

show that the average of a series given by: 2+12+30+56+90+...+nth term. is given by: (4n-1)(n+1)/3, where n is the number of terms.

#Algebra

Note by Kelvin Eden
6 months, 2 weeks ago

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Comments

Nice series! I really enjoyed working this out. Not sure if this is the standard approach, but this is how I did it.

Each term in the series can be represented by the following expression, where tt is the term's placement in the series (in ttth place, starting from 11):

(2t)(2t−1)=4t2−2t(2t)(2t - 1) = 4t^2 - 2t

This is because one will notice that the first term is 1×21\times2, the second is 3×43\times4, the third 5×65\times6 and so on. To find the average of this series up to a number nn, we need to sum every term together and divide by nn. This can be expressed as:

1n∑t=1n4t2−2t\dfrac{1}{n}\displaystyle\sum_{t = 1}^{n}4t^2-2t

The simplification of this expression is as follows:

1n(4∑t=1nt2−2∑t=1nt)\displaystyle\dfrac{1}{n}\left(4\sum_{t = 1}^{n}t^2-2\sum_{t = 1}^{n}t\right)

1n(4n(n+1)(2n+1)6−2n(n+1)2)\dfrac{1}{n}\left(\dfrac{4n(n+1)(2n+1)}{6}-\dfrac{2n(n+1)}{2}\right)

2(n+1)(2n+1)3−(n+1)\dfrac{2(n+1)(2n+1)}{3}-(n+1)

(n+1)(23(2n+1)−1)(n+1)\left(\dfrac{2}{3}(2n+1)-1\right)

(n+1)(43n−13)(n+1)\left(\dfrac{4}{3}n - \dfrac{1}{3}\right)

(n+1)(4n−13)(n+1)\left(\dfrac{4n - 1}{3}\right)

4(n−1)(n+1)3\boxed{\dfrac{4(n-1)(n+1)}{3}}

David Stiff - 6 months, 2 weeks ago

I tried to solve this problem with a different approach, but this one is simpler and more clear... perfect👏👏

Kelvin Eden - 6 months, 2 weeks ago

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Thanks!

David Stiff - 6 months, 2 weeks ago
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