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Question

Evaluate the series. Do not convert your answer to a decimal. 22. sigma_k=1^17 (In 2 - kpi) 23. sigma_j=1^10 pi/180 (2)^j-1 24. Find ihe sum of all of the 3-digit positive integers. 25. Find the sum of every integer from 1 to 1000 that is not a perfcct square.

Answer 1

#25 to get you started...

find the sum of every integer from 1 to 1000 that is not a perfect square

find the sum of all the numbers from 1 to 1000 and then subtract the sum of all the squares from 1 to 1000

the sum of the first n consecutive numbers is found by using the formula n(n+1)/2

1000(1001)/2=500(1001)=500,500

you could have also done the following:

1+2+3+4+5+...+995+996+997+998+999+1000

1+1000=1001

2+999=1001

3+998=1001

and so on

there are 500 of these 1001 sums

500*1001=500,500 again

there is a formula for the sum of all the squares from 1 to 1000

[n(n+1)(2n+1)]/6

you have to determine what n is by trial and error

or take the square root of 1000 which is 31.6 so n=31

30²=900, 31²=961, 32²=1024, so n must be 31 because 32² is greater than 1000

[31(31+1)(2*31+1)]/6=[31(32)(63)]/6=(62,496)/6=10,416

500,500-10,416=490,084

 

#24) you can find the sum of the numbers from 1 to 999 and subtract the sum of the numbers from 1 to 99 or you can find the sum using an arithmetic sequence solution... 999(1000)/2=999(500)=499,500 99(100)/2=99(50)=4950 499,500-4950=494,550 or... there are 900 3-digit numbers from 100 to 999 S(900)=(900/2)(2*100+[900-1][1]) S(900)=(450)(200+899) S(900)=(450)(1099) S(900)=494,550