1+2+3+4+5+6+7+8+9......+99999999999+100000000000

🎓 學業台小學及中學教育

無聊

2006-10-23 12:02 am

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回答 (7)

Kwanie

2006-10-23 12:08 am

✔ 最佳答案

1+2+3+4+5+6+7+8+9......+99999999999+100000000000
=[(1+100000000000) * 100000000000]/2
=100000000001 * 50000000000
=5000000000050000000000

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※千鳥※

2006-10-23 1:44 am

((1+100000000000)*100000000000)/2
=5000000000050000000000

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文詩

2006-10-23 12:52 am

5000000000050000000000

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用戶173561

2006-10-23 12:19 am

a=1
d=100000000000
T(n)=1+(n-1)100000000000
T(n)=1+100000000000n-100000000000
T(n)=-99999999999+100000000000n
-100000000000n=-99999999999
n=0.99999999999
S(n)=n/2[2a+(n-1)d]
S(0.99999999999)=0.99999999999/2[2*1+(0.99999999999-1)100000000000]
S(0.99999999999)=0.499999999995[2+(-0.00000000001)100000000000]
S(0.99999999999)=0.499999999995

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祖葉

2006-10-23 12:06 am

(1+100000000000)x 100000000000 /2

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