document.write( "Question 860335: use mathematical induction to prove that $\"%281%5E2%29%2B%282%5E2%29%2B%283%5E2%29\"$ +...+ $\"%282%5En%29=n%5E%28k%2B1%29+-2\"$ \n" ); document.write( "

Algebra.Com's Answer #518395 by KMST(5328) $\"\"$ $\"About$

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You want to prove that
\n" ); document.write( " $\"2%5E1%2B2%5E2%2B2%5E3%2B%22...%22%2B2%5En=2%5E%28n%2B1%29-2\"$
\n" ); document.write( "or, using fancier symbols, that
\n" ); document.write( " $\"sum%282%5Ep%2Cp=1%2Cp=n%29=2%5E%28n%2B1%29-2\"$ .
\n" ); document.write( "You need to prove that
\n" ); document.write( "a) it is true for $\"n=1\"$ , and
\n" ); document.write( "b) if it is true for $\"n=k\"$ , then it will be true for $\"n=k%2B1\"$ .
\n" ); document.write( "
\n" ); document.write( "For $\"n=1\"$ the \"sum\" has only one term:
\n" ); document.write( " $\"sum%282%5Ep%2Cp=1%2Cp=n%29=sum%282%5Ep%2Cp=1%2Cp=1%29=2%5E1=2\"$
\n" ); document.write( "and it is indeed equal to
\n" ); document.write( " $\"2%5E%28n%2B1%29-2=2%5E%281%2B1%29-2=2%5E2-2=4-2=2\"$
\n" ); document.write( "
\n" ); document.write( "If the formula $\"2%5E1%2B2%5E2%2B2%5E3%2B%22...%22%2B2%5En=2%5E%28n%2B1%29-2\"$ is true for $\"n=k\"$ ,
\n" ); document.write( " $\"2%5E1%2B2%5E2%2B2%5E3%2B%22...%22%2B2%5Ek=2%5E%28k%2B1%29-2\"$ .
\n" ); document.write( "Adding one more term, we find
\n" ); document.write( "

,
\n" ); document.write( "which shows that the formula is true for $\"n=k%2B1\"$ . \n" ); document.write( "

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