Find the sum of first 1000 positive inyegers
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The answer is 500500
plz mark as branliest if correct
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The sum you are looking for is:
2+4+6+8+...+2000, but taking 2 as a common factor this is the same as:
2(1+2+3+4+...+1000)
Now we can use the formula for the sum:
1000∑n=1n=1000⋅10012=500⋅1001=500500
Using this now we have:
2(1+2+3+4+...+1000)=2⋅1000∑n=1n=1000⋅10012=
=2⋅500500=1001000
2+4+6+8+...+2000, but taking 2 as a common factor this is the same as:
2(1+2+3+4+...+1000)
Now we can use the formula for the sum:
1000∑n=1n=1000⋅10012=500⋅1001=500500
Using this now we have:
2(1+2+3+4+...+1000)=2⋅1000∑n=1n=1000⋅10012=
=2⋅500500=1001000
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