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klb15
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PS-Sum of even integers.jpg 10 Oct 2008, 06:47
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kabaddi
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PS-Sum of even integers.jpg 10 Oct 2008, 14:40
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7550.
explanation-
2+4+6+...+100 = 2550.
sum of all even integers from 102 to 200 is
102+104+106+...+200
=(100+2)+(100+4)+(100+6)+....+(100+100)
=(2+4+6+....+100)+(50*100)
=2550+5000
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LiveStronger
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PS-Sum of even integers.jpg 10 Oct 2008, 14:48
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7550

102 + 104 + 106 + .... 200
100 + 2 + 100 + 4 + 100 + 6 ..... 100+100
50*100 + (2+4+6 ...100)

= 5000 + 2550 = 7550
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FN
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PS-Sum of even integers.jpg 10 Oct 2008, 14:54
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first figure how how many integers there are between 102-200

200=102+(n-1)(2) => n=50

sum= 50/2 (2*102+49*2)=7550
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Greenberg
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PS-Sum of even integers.jpg 10 Oct 2008, 15:43
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lets see an easy one

2+4+6+8+10 = 30

2(1+2+3+4+5) = 30

we know that 1+2+3+4+5 = 5*6/2 (the last one*last one+1)/2

so 2*(5*6/2) = 30

solve

2*(100*101)/2-2*(50*51)/2 = 7550

:)



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