document.write( "Question 892087: Find the sum of
\n" ); document.write( "(1+2+3....+x) + (2+3+4....+x) + (3+4+5....+x) + ...................... till n terms where n\n" ); document.write( "

Algebra.Com's Answer #540308 by robertb(5830) $\"\"$ $\"About$

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Arrange the sums in this manner:\r
\n" ); document.write( "\n" ); document.write( "1 + 2 + 3 + 4 + ... + n + (n+1) + ... + x
\n" ); document.write( " 2 + 3 + 4 + ... + n + (n+1) + ... + x
\n" ); document.write( " 3 + 4 + ... + n + (n+1) + ... + x
\n" ); document.write( " 4 + ... + n + (n+1) + ... + x
\n" ); document.write( " ...
\n" ); document.write( " (n-1)+n + (n+1) + ... + x
\n" ); document.write( " n + (n+1) + ... + x
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\n" ); document.write( "The triangular portion of the sums can be expressed as
\n" ); document.write( "1 + 2*2 + 3*3 + 4*4 +... + (n-1)*(n-1) + n*n, or
\n" ); document.write( " $\"1%2B2%5E2+%2B+3%5E2+%2B4%5E2\"$ +...+ $\"%28n-1%29%5E2+%2Bn%5E2+=+%28n%2A%28n%2B1%29%2A%282n%2B1%29%29%2F6\"$
\n" ); document.write( "Now the sum (n+1)+...+x is added n times. Hence
\n" ); document.write( "the sum $\"n%2A%28%28x-n%29%2F2%29%2A%28n%2B1%2Bx%29\"$ , by using the sum of an arithmetic sequence.\r
\n" ); document.write( "\n" ); document.write( "Therefore, (1+2+3....+x) + (2+3+4....+x) + (3+4+5....+x) + .... till n terms where n < x, is
\n" ); document.write( " $\"%28n%2A%28n%2B1%29%2A%282n%2B1%29%29%2F6+%2B+%28n%2A%28x-n%29%2A%28n%2B1%2Bx%29%29%2F2\"$
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