+23248 One way to do this that will work for other questions of this type (that is, for other number of terms and for other sum) is the following (but it won't by a 'pretty' answer:
Since we want 50 terms, add the first 50 whole numbers: 1 + 2 + 3 + ... + 48 + 49 + 50.
There is a formula that allows you do add these quickly: n = number of terms, f = first term, l = last term:
Sum = n(f+l)/2
Since n = 50, first = 1, and last = 50: Sum = 50(1+50)/2 ---> Sum = 50(51)/2 ---> Sum =1275
But the sum is 1275 and we want a sum of 2000, we need to 'scale up' every number.
How much? Since we want 2000 and not 1275, the scale factor will be 2000/1275 or 80/51.
Let's multiply each term by 80/51; this will then multiply the answer by 80/51 (and 80/51 x 1275 = 2000).
The terms are 1(80/51) + 2(80/51) + 3(80/51) + ... + 48(80/51) + 49(80/51) + 50(80/51)
or: 80/51 + 160/51 + 240/51 + ... + 3840/51 + 3920/51 + 4000/51
+23248 One way to do this that will work for other questions of this type (that is, for other number of terms and for other sum) is the following (but it won't by a 'pretty' answer:
Since we want 50 terms, add the first 50 whole numbers: 1 + 2 + 3 + ... + 48 + 49 + 50.
There is a formula that allows you do add these quickly: n = number of terms, f = first term, l = last term:
Sum = n(f+l)/2
Since n = 50, first = 1, and last = 50: Sum = 50(1+50)/2 ---> Sum = 50(51)/2 ---> Sum =1275
But the sum is 1275 and we want a sum of 2000, we need to 'scale up' every number.
How much? Since we want 2000 and not 1275, the scale factor will be 2000/1275 or 80/51.
Let's multiply each term by 80/51; this will then multiply the answer by 80/51 (and 80/51 x 1275 = 2000).
The terms are 1(80/51) + 2(80/51) + 3(80/51) + ... + 48(80/51) + 49(80/51) + 50(80/51)
or: 80/51 + 160/51 + 240/51 + ... + 3840/51 + 3920/51 + 4000/51
