S suf Jun 2010 3 0 Jun 9, 2010 #1 This is question 4 on June 2008 Further Pure 1 paper OCR (downloadable for OCR website) OCR > Qualifications > By type > AS/A Level GCE (current) > Mathematics > Mathematics > All documents The question goes: The Matrix A is given by A = 3, 1, 0, 1, prove by induction that for n>1 A^n = 3^n, 0.5(3^n -1), 0, 1, I just dont get how 3^k + 0.5(3^k -1) becomes 0.5(3^k+1 -1) Thanks

This is question 4 on June 2008 Further Pure 1 paper OCR (downloadable for OCR website) OCR > Qualifications > By type > AS/A Level GCE (current) > Mathematics > Mathematics > All documents The question goes: The Matrix A is given by A = 3, 1, 0, 1, prove by induction that for n>1 A^n = 3^n, 0.5(3^n -1), 0, 1, I just dont get how 3^k + 0.5(3^k -1) becomes 0.5(3^k+1 -1) Thanks

slider142 May 2009 72 21 Brooklyn, NY Jun 9, 2010 #2 suf said: ... I just dont get how 3^k + 0.5(3^k -1) becomes 0.5(3^k+1 -1) Thanks Click to expand... Simplify the expression: \(\displaystyle 3^k + \frac{1}{2} 3^k - \frac{1}{2}\)

suf said: ... I just dont get how 3^k + 0.5(3^k -1) becomes 0.5(3^k+1 -1) Thanks Click to expand... Simplify the expression: \(\displaystyle 3^k + \frac{1}{2} 3^k - \frac{1}{2}\)

Prove It MHF Helper Aug 2008 12,897 5,001 Jun 10, 2010 #4 slider142 said: Simplify the expression: \(\displaystyle 3^k + \frac{1}{2} 3^k - \frac{1}{2}\) Click to expand... \(\displaystyle 3^k + \frac{1}{2}\cdot 3^k - \frac{1}{2} = \left(1 + \frac{1}{2}\right)3^k - \frac{1}{2}\) \(\displaystyle = \frac{3}{2}\cdot 3^k - \frac{1}{2}\) \(\displaystyle = \frac{1}{2}\cdot 3^{k + 1} - \frac{1}{2}\) \(\displaystyle = \frac{1}{2}(3^k - 1)\). Reactions: suf

slider142 said: Simplify the expression: \(\displaystyle 3^k + \frac{1}{2} 3^k - \frac{1}{2}\) Click to expand... \(\displaystyle 3^k + \frac{1}{2}\cdot 3^k - \frac{1}{2} = \left(1 + \frac{1}{2}\right)3^k - \frac{1}{2}\) \(\displaystyle = \frac{3}{2}\cdot 3^k - \frac{1}{2}\) \(\displaystyle = \frac{1}{2}\cdot 3^{k + 1} - \frac{1}{2}\) \(\displaystyle = \frac{1}{2}(3^k - 1)\).