Let f(x) equal the sum of all of the integers from 1 to x, inclusive,

Question

Let f(x) equal the sum of all of the integers from 1 to x, inclusive, where x ≥ 1. If \(g=\frac{f(1)}{1^2} + \frac{f(2)}{2^2} + \frac{f(3)}{3^2} + ...+ \frac{f(10)}{10^2}\), which of the following is true of g ?

(A) 0 ≤ g < 5

(B) 5 ≤ g < 10

(C) 10 ≤ g < 15

(D) 15 ≤ g < 20

(E) 20 ≤ g < 25

(A) 0 ≤ g < 5

(B) 5 ≤ g < 10

(C) 10 ≤ g < 15

(D) 15 ≤ g < 20

(E) 20 ≤ g < 25

BrushMyQuant · Answer

Given that f(x) equal the sum of all of the integers from 1 to x, inclusive, where x ≥ 1 and \(g=\frac{f(1)}{1^2} + \frac{f(2)}{2^2} + \frac{f(3)}{3^2} + ...+ \frac{f(10)}{10^2}\) and we need to find the approximate value of g

To find f(1) we need to compare what is in the bracket in f(1) and what is in the bracket in f(x)

=> to find the value of f(1) we need to replace x with 1 in f(x)
=> f(1) = Sum of all integers from 1 to 1 = 1
=> f(2) = 1+2 = 3
=> f(3) = 1+2+3 = 6
.
.
.
=> f(10) = 1+2+3...+10 = \(\frac{10*11}{2}\) = 55

=> \(g=\frac{f(1)}{1^2} + \frac{f(2)}{2^2} + \frac{f(3)}{3^2} + ...+ \frac{f(10)}{10^2}\)
=> g = \(\frac{1}{1} + \frac{3}{4} + \frac{6}{9} + \frac{10}{16} +... + \frac{55}{100} \)

Now each term apart from the first one is greater than 0.5 and less than 1
=> g > 1 + 0.5*9 = 1+4.5 = 5.5
g < 1 + 9*1 = 10
=> 5 ≤ g < 10

So, Answer will be B
Hope it helps!

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JerryAtDreamScore · Answer

Breaking Down the Info:

f(1) = 1

f(10) = 1 + 2 + 3 + ... + 9 + 10 = 11*5 = 55.

There are 10 terms in g, with the biggest term being 1, and the smallest term being  \frac{f(10)}{10^2} = 0.55 . Since all terms fall in this range, we can see the total is at most 10, and at least 5.5. Therefore we can choose B.

Answer: B

IanStewart · Answer

If you list the first n positive integers, you have an equally spaced list. So the average of the list will be (n+1)/2, the average of the smallest and largest terms. Since sum = (number of terms)*average, and since we have n integers, the sum of the first n positive integers is always n(n+1)/2 = (n^2 + n)/2.

So in this question, f(n) = (n^2 + n)/2, and since each term in the sum is f(n)/n^2, each term is equal to   / n^2 = (n^2 + n)/2n^2 = 1/2 + 1/(2n). So each term is clearly bigger than 1/2, since each term is equal to 1/2 plus something positive, and each term after the first is clearly less than 1, because we're adding 1/2 to 1/(2n) which is less than 1/2 if n > 1 (and is exactly 1/2 when n = 1). We have 10 terms, so the sum will be between 10(1/2) and 10*1, or between 5 and 10.

This method might be overkill for this particular question, since if you work out a few terms you can make a reasonable guess at what is going on, but this method would adapt to any situation, e.g. if the question asked about a longer sequence, or about one that doesn't start at n = 1. The other solutions currently in this thread assume the terms get progressively smaller as you get deeper into this sum, which really needs some justification; that turns out to be true here, but there are lots of natural sequences where terms in the middle are larger than the terms at the start or at the end.

bumpbot · Answer

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Let f(x) equal the sum of all of the integers from 1 to x, inclusive,

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Let f(x) equal the sum of all of the integers from 1 to x, inclusive,

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