document.write( "Question 137056: Is there a short way to do the SUM of a sequence of sums: (1) + (1+2) + (1+2+3) + (1+2+3+4) + .... (1+2+3+4....+100). I know how to do the sum of the sequence (i.e. sum of 100= 100/2(100 + 1)) and I can get the answer, I am just trying to see if there is an easier way then adding each number up. Like, sum for 100 is 5050; sum of 99 is sum of 100 minus 100 or 4050; but I can't find a shortcut and I can't find a formula anywhere for the sum of a series of sums. Can someone help? This is actually not for a class. \n" ); document.write( "
Algebra.Com's Answer #100303 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
Given: The first term of every summation in the series is \"a\".
\n" ); document.write( "Given: The number of numbers in each summation is 1 greater than the previous term.
\n" ); document.write( "Given: There are \"n\" terms in the series.\r
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\n" ); document.write( "\n" ); document.write( "The first term of your series must be \"na\", because every term in the series will have an \"a\" in it.\r
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\n" ); document.write( "\n" ); document.write( "The second term of your series must be \"%28n-1%29%28a%2B1%29\", because all but the first term will have an \"a%2B1\" in it. This expands to: \"na%2Bn-a-1\"\r
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\n" ); document.write( "\n" ); document.write( "The third term of your series must be \"%28n-2%29%28a%2B2%29\", similar logic, and this expands to \"na%2B2n-2a%2B4\"\r
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\n" ); document.write( "\n" ); document.write( "This gives us the idea that the i-th term must be \"%28n-%28i-1%29%29%28a%2B%28i-1%29%29\", which expands to \"na%2B%28i-1%29n-%28i-1%29a-%28i-1%29%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Notice that every term has an \"na\" term, so there must be \"n\" times \"na\" in your sum. The first term of the shortcut formula is then \"n%5E2a\"\r
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\n" ); document.write( "\n" ); document.write( "If you sum all of then \"n\" terms, you find that the coefficients can be expressed as \"sum%28%28i-1%29%2Ci=1%2Cn%29=sum%28i%2Ci=0%2Cn%29=sum%28i%2Ci=1%2Cn-1%29\". Since \"sum%28i%2Ci=1%2Cn%29=%28n%28n%2B1%29%29%2F2\", \"sum%28i%2Ci=1%2Cn-1%29=%28n%28n-1%29%29%2F2\". Hence, the second term of the shortcut formula must be \"%28n%5E2%28n-1%29%29%2F2\".\r
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\n" ); document.write( "\n" ); document.write( "Similarly, the sum of all the \"a\" terms must be \"-%28na%28n-1%29%29%2F2\".\r
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\n" ); document.write( "\n" ); document.write( "Finally, the last term is because the sum of squares is \"sum%28i%5E2%2Ci=1%2Cn%29=%28n%28n%2B1%29%282n%2B1%29%29%2F6\".\r
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\n" ); document.write( "\n" ); document.write( "Putting it all together:\r
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\n" ); document.write( "\n" ); document.write( "I won't go into the details of the derivation, but if you have a common difference other than 1, call it \"d\", the formula becomes:\r
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