# Showing that the formula is correct.

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• Apr 7th 2011, 09:34 AM
Rosie
Showing that the formula is correct.
Okay this is a series and sequence equation.

I do not know how to prove that the following equation can be wrote in that way. The question is:

20;10;5; .... is a G.S.

Show that the n^th term of the sequence is given by the formula:
Tn= 5.2^(3-n)

Please help me.
• Apr 7th 2011, 09:41 AM
NOX Andrew
The sequence can be written as

$5 \cdot 4, 5 \cdot 2, 5 \cdot 1, ... = 5 \cdot 2^2, 5 \cdot 2^1, 5 \cdot 2^0, ...$

As you can see, every term is 5 times a power of 2. To be precise, the nth term is 5 times the (3 - n)th power of 2, or

$T_n = 5 \cdot 2^{3-n}$
• Apr 7th 2011, 10:18 AM
Rosie
Still do not understand, why did the ^(n-1) go to ^(3-n), how did see that the sequence can be wrote diffrent?
• Apr 7th 2011, 11:23 AM
NOX Andrew
In answer to how I realized the sequence could be written as

$5 \cdot 2^2$, $5 \cdot 2^1$, $5 \cdot 2^0$, ...

it's a matter of pattern recognition. I noticed any given term was one-half the previous term, so I knew powers of 2 were involved ( $\dfrac{1}{2} = 2^{-1}$). I also noticed the first three terms were multiples of 5, so I knew 5 was involved. Putting it all together, I re-wrote each term as 5 times a power of 2.

In answer to why the exponent is $3 - n$, consider the exponents of the first three terms. For the 1st term, the exponent is 2. For the 2nd term, the exponent is 1. For the 3rd term, the exponent is 0. In other words, the exponent decreases in successive terms. Thus, $-n$ is appropriate because $-n$ decreases as $n$ increases.

The $3$ in the exponent is necessary to "align" the terms and the exponents. If it weren't there, then the exponent of the 1st term would be -1. However, the correct exponent of the 1st term is 2, so ask yourself what could you add to -1 to produce 2. The answer is 3. Thus, the correct exponent is $-n + 3 = 3 - n$.