Daily Question Contest 2

=1/a^2

Hi Mathhemathh,

I am really pleased that you have become a member.  Welcome to our web2.0calc.com  forum.      

This is the sum of a GP over the sum of another GP

Both have a common ratio of a  and both have 7 terms      r=a   and n=7

\(\mbox{The sum of a GP } = \frac{T_1(r^n-1)}{r-1}    \)

\(answer \\ = \left( \frac{a(a^7-1)}{(a-1)} \right ) \div \left(\frac{a^3 (a^7-1)}{(a-1)} \right)\\ = \left( \frac{a(a^7-1)}{(a-1)} \right ) \times \left(\frac{(a-1)}{a^3 (a^7-1)} \right)\\ =\frac{1}{a^2}\)

Mathhemathh  ...   What does this mean?

3.   The 3 winners will be in the next question

OH I get it, you mean your will give them credit for their answers.    

OK, glad you get it! 

LINK TO THE FIRST DAILY QUESTION CONTEST:

http://web2.0calc.com/questions/daily-question-contest-1

LINK TO IMAGINARY NUMBERS 1:

http://web2.0calc.com/questions/imaginary-numbers-1

\(\begin{array}{rcll} \dfrac{a+a^2+a^3+a^4+a^5+a^6+a^7}{a^3+a^4+a^5+a^6+a^7+a^8+a^9} &=& \dfrac{a\cdot(1+a^1+a^2+a^3+a^4+a^5+a^6)}{a^3\cdot(1+a^1+a^2+a^3+a^4+a^5+a^6)} \\\\ &=& \dfrac{a}{a^3} \\\\ &=& \dfrac{1}{a^2} \\\\ \end{array}\)