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

(a) 3(3k - 1)/2        (b) 3(3k - 1)/2 + 3k+1

Halp!!!!!!!

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   Feb 18, 2016
edited by CPhill  Feb 18, 2016
edited by CPhill  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
edited by CPhill  Feb 18, 2016
edited by CPhill  Feb 18, 2016