Show that replacing k with k + 1 in problem (a) yields and expression equivalent to (b):

(a) 3(3^{k} - 1)/2 (b) 3(3^{k} - 1)/2 + 3^{k+1}

Halp!!!!!!!

Shades
Feb 18, 2016

#1**+10 **

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

CPhill
Feb 18, 2016

#1**+10 **

Best Answer

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

CPhill
Feb 18, 2016