Use mathematical induction to prove that for all positive integers n:
1. a+ar+ar^2+...+ar^(n-1)= a(r^n-1)/r-1, provided r does not equal to 1
2. a+(a+d)+(a+2d)...+(a+(n-1)d)=1/2 x n(2a+(n-1)d)
I just don't know what to do when there are more than one different... what's it called... letters (?) in the question.
1. a+ar+ar^2+...+ar^(n-1)= a(r^n-1)/r-1, provided r does not equal to 1
2. a+(a+d)+(a+2d)...+(a+(n-1)d)=1/2 x n(2a+(n-1)d)
I just don't know what to do when there are more than one different... what's it called... letters (?) in the question.