document.write( "Question 736005: Prove using simple induction that, for each integer n >= 1
\n" ); document.write( "3 + 3^2+ 3^3+...+ 3^n = (3^(n+1)-3)/2
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Algebra.Com's Answer #449630 by stanbon(75887) $\"\"$ $\"About$

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Prove using simple induction that, for each integer n >= 1
\n" ); document.write( "3 + 3^2+ 3^3+...+ 3^n = (3^(n+1)-3)/2
\n" ); document.write( "----
\n" ); document.write( "Show it is true for n = 1
\n" ); document.write( "3 = (3^(1+1)-3)/2 = (3^2 - 3)/2 = 6/2 = 3
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\n" ); document.write( "Assume it is true for n = k:
\n" ); document.write( "3 + 3^2 +...+3^k = (3^(k+1) -3)/2
\n" ); document.write( "----
\n" ); document.write( "Show it must be true for n = k+1
\n" ); document.write( "[3 + 3^2+...+3^k] + 3^(k+1)
\n" ); document.write( "---
\n" ); document.write( "= (3^(k+1) -3)/2 + 3^(k+1)
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\n" ); document.write( "= [3^(k+1) - 3) + 2*3^(k+1)]/2
\n" ); document.write( "----
\n" ); document.write( "= [3*3^(k+1) - 3]/2
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\n" ); document.write( "= [3^((k+1)+1) - 3]/2
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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