document.write( "Question 1184007: prove by mathematical induction:
\n" ); document.write( "1^5 + 2^5 + 3^5 + ... + n^5 = (1/12)n^2(n+1)^2(2n^2 + 2n -1)
\n" ); document.write( "

Algebra.Com's Answer #852303 by ikleyn(52832)\"\" \"About 
You can put this solution on YOUR website!
.
\n" ); document.write( "Prove by induction that for all n >= 1
\n" ); document.write( "1^ 5 +2^ 5 +3^ 5 +^ ...+ n ^ 5 = [n ^ 2 * (n + 1) ^ 2 * (2n ^ 2 + 2n - 1)] / 12 .
\n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\r\n" );
document.write( "(a) The base of induction: n = 1.\r\n" );
document.write( "\r\n" );
document.write( "    Then the sum is one single term  \"1%5E5\",  which is 1.\r\n" );
document.write( "\r\n" );
document.write( "    The formula (*) at n = 1 gives \r\n" );
document.write( "\r\n" );
document.write( "        \"%281%2F12%29%2A%281%5E2%2A%281%2B1%29%5E2%2A%282%2A1%5E2+%2B+2%2A1-1%29%29\" = \"%281%2F12%29%2A%282%5E2%2A%282%2B2-1%29%29\" = \"%281%2F12%29%2A4%2A3\" = \"%281%2F12%29%2A12\" = 1,  \r\n" );
document.write( "\r\n" );
document.write( "    so the base of induction is established.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "(b)  The step of induction.\r\n" );
document.write( "\r\n" );
document.write( "     We assume that for some integer k >= 1 this formula is valid\r\n" );
document.write( "\r\n" );
document.write( "         1^5 + 2^5 + 3^5 + . . . + k^5 = \"%281%2F12%29%2Ak%5E2%2A%28k%2B1%29%5E2%2A%282k%5E2%2B2k-1%29\".     (1)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "     We want to prove that then the formula is valid for the next integer number k+1, too:\r\n" );
document.write( "\r\n" );
document.write( "         1^5 + 2^5 + 3^5 + . . . + k^5 + (k+1)^5 = \"%281%2F12%29%2A%28k%2B1%29%5E2%2A%28k%2B2%29%5E2%2A%282%28k%2B1%29%5E2%2B2%28k%2B1%29-1%29\".     (2)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "     At this point, the proof of the formula (2) is started.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "     In the left side of (2), we replace the sum of the first k addends by the right side expression (1).\r\n" );
document.write( "\r\n" );
document.write( "     Thus we want to prove\r\n" );
document.write( "\r\n" );
document.write( "         \"%281%2F12%29%2Ak%5E2%2A%28k%2B1%29%5E2%2A%282k%5E2%2B2k-1%29\" + \"%28k%2B1%29%5E5\" = \"%281%2F12%29%2A%28k%2B1%29%5E2%2A%28k%2B2%29%5E2%2A%282%28k%2B1%29%5E2%2B2%28k%2B1%29-1%29\".     (3)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "     Let's transform left side of (3). We factor it, taking the common factor \"%28k%2B1%29%5E2\" out of parentheses.\r\n" );
document.write( "\r\n" );
document.write( "     Then left side of (3) takes the form\r\n" );
document.write( "\r\n" );
document.write( "           \"%28k%2B1%29%5E2+%2A+%281%2F12%29%2Ak%5E2%2A%28%282k%5E2%2B2k-1%29%2B12%28k%2B1%29%5E3%29\" = \r\n" );
document.write( "\r\n" );
document.write( "         =  = \r\n" );
document.write( "\r\n" );
document.write( "         = \"%281%2F12%29%2A%28k%2B1%29%5E2%2A%282k%5E4+%2B+2k%5E3+%2B+35k%5E2+%2B+36k+%2B+12%29\".    (4)\r\n" );
document.write( "\r\n" );
document.write( "    \r\n" );
document.write( "     Now, I used an online calculator to factor an expression in the internal parentheses,\r\n" );
document.write( "     and the calculator produced this decomposition\r\n" );
document.write( "\r\n" );
document.write( "         \"2k%5E4+%2B+2k%5E3+%2B+35k%5E2+%2B+36k+%2B+12\" = \"%28k%2B2%29%5E2%2A%282k%5E2%2B6k%2B3%29\".     (5)\r\n" );
document.write( "\r\n" );
document.write( "     ( the link to the calculator is  https://www.pocketmath.net , the mode is \"Factor\" )  )\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "     This factorization can be continued this way\r\n" );
document.write( "     \r\n" );
document.write( "         \"2k%5E4+%2B+2k%5E3+%2B+35k%5E2+%2B+36k+%2B+12\" = \"%28k%2B2%29%5E2%2A%282k%5E2%2B6k%2B3%29\" = \"%28k%2B2%29%5E2%2A%282%28k%2B1%29%5E2%2B2%28k%2B1%29-1%29\".    (6)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "     Now, combining all pieces (4), (5) and (6) in one whole block, we have\r\n" );
document.write( "\r\n" );
document.write( "         \"%281%2F12%29%2Ak%5E2%2A%28k%2B1%29%5E2%2A%282k%5E2%2B2k-1%29\" + \"%28k%2B1%29%5E5\" = \"%281%2F12%29%2A%28k%2B1%29%5E2%2A%28k%2B2%29%5E2%2A%282%28k%2B1%29%5E2%2B2%28k%2B1%29-1%29\".   (7)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "     It is the same as (identical to) formula (3).  Thus formula (3) is proven.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "(3)  Due to the principle of the mathematical induction, it means that formula \r\n" );
document.write( "\r\n" );
document.write( "         \"1%5E5+%2B+2%5E5+%2B+3%5E5+%2B+ellipsis+%2B+n%5E5\" = \"%281%2F12%29%2An%5E2%2A%28n%2B1%29%5E2%2A%282n%5E2%2B2n-1%29\".     \r\n" );
document.write( "\r\n" );
document.write( "     is proved for all integer n >= 1.\r\n" );
document.write( "
\r
\n" ); document.write( "\n" ); document.write( "Solved.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );