Shades....I'm going to show that (b) can be made to look like (a) where k is replaced by k + 1
All you would need to do is to reverse the steps [if you wanted to ]
3(3^k - 1)/2 + [ 3^(k+1)] =
Note 3^(k +1) = 3*3^k........also....split the first term into two fractions
[3*3^k ]/ 2 - 3/2 + [3*3^k] =
Get a common denominator in the third fraction by multiplying by 2 on top/bottom
[3*3^k ]/ 2 - 3/2 + 2*[3*3^k]/2
Put everything over 2
(3*3^k - 3 + 2*[3*3^k] ) / 2 =
[3*3^k - 3 + 6*3^k] / 2 =
Combine like terms
[9*3^k - 3] / 2 =
Factor out a 3
3 [3*3^k -1] / 2 =
Note again, 3*3^k = 3^(k + 1)
3 [ 3^(k + 1) - 1] / 2
