AAPL wrote:Manhattan Prep
If \(x\) is a positive integer greater than 1, what is the sum of the multiples of \(x\) from \(x\) to \(x^2\), inclusive?
A. \(x(x+1)(x-1)\)
B. \(\frac{x^2(x+1)}{2}\)
C. \(x^2(x-1)\)
D. \(\frac{x^3+2x}{2}\)
E. \(x(x-1)^2\)
OA B
Since we know that x is a positive integer greater than 1, we can let x = 2. Thus, we need to determine the sum of multiples of 2 from 2 to 4 inclusive. We see that the sum of the multiples of 2 is 2 + 4 = 6. Now we need to determine which of the answer choices is equivalent to 6:
A) x(x + 1)(x - 1) = 2(3)(1) = 6....YES
B) x^2(x + 1)/2 = 4(3)/2 = 6....YES
C) x^2(x - 1) = 4(1) = 4....NO
D) (x^3 + 2x)/2 = (8 + 4)/2 = 6....YES
E) x(x - 1)^2 = 2(1)^2 = 2....NO
To decide among answer choices A, B, and D, we can let x = 3. The multiples of 3 between 3 and 9 inclusive are 3, 6, and 9. We have 3 + 6 + 9 = 18. Let's plug x = 3 into answer choices A, B, and D and see which one(s) produce 18:
A) x(x + 1)(x - 1) = 3(4)(2) = 24....NO
B) x^2(x + 1)/2 =9(4)/2 = 18....YES
D) (x^3 + 2x)/2 = (27 + 6)/2 = 16.5....NO
Alternate Solution:
The multiples of x between x and x^2 inclusive are x, 2x, 3x, ... , (x - 1)x, (x)(x) = x^2.
This sum is: x + 2x + 3x + ... + (x - 1)x + x^2.
Let's factor out the x: x(1 + 2 + 3 + ... + x).
Note that the expression in parentheses is simply the sum of the first consecutive x integers, so it equals [x(x+1)]/2. Therefore, x + 2x + 3x + ... + (x - 1)x + x^2 = x*[x(x + 1)]/2 = x^2(x + 1)/2.
Answer: B
