Question 550929
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I'll do 4 of them.  You can do the other 4.  They are similar to these.

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1. 42 is on top of the symbol, i=1 is on bottom and 1 on right side. 
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{{{sum(1,i=1,42)}}} = 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 = 42
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2. 5 on top, n on right, n=1 on bottom. the answer is 15 but how?
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{{{sum(n,1=1,5)}}} = 1+2+3+4+5 = 15 We didn't need a formula but we can use it:
              {{{n(n+1)/2}}} = {{{5(5+1)/2}}} = {{{((5)(6))/2}}} = 15
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4. 20 on top, k on right, k=1 on bottom
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{{{sum(k,k=1,20)}}} = 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20

We use the formula  {{{n(n+1)/2}}} = {{{20(20+1)/2}}} = {{{((20)(21))/2}}} = 210
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8. 35 on top, k^2 on right, k=1 on bottom
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{{{sum(k^2,i=1,35)}}} = 1²+2²+3²+4²+5²+6²+7²+8²+9²+10²+11²+12²+13²+14²+15²+16²+17²+18²+19²+20²+21²+22²+23²+24²+25²+26²+27²+28²+29²+30²+31²+32²+33²+34²+35²
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We use the formula {{{n(n+1)(2n+1)/6}}} = {{{35(35+1)(2*35+1)/6}}} = {{{((35)(36)(70+1))/6}}} = {{{((35)(36)(71))/6}}} = 14910.

Edwin</pre>