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If \(x\) is a positive integer greater than 1, what is the

Expert replies
by AAPL » Fri Jul 19, 2019 5:38 am

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Answers

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B

C

D

E

Stats

Difficulty—

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
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Source: — Problem Solving |

by Piyush_Pandey » Fri Jul 19, 2019 7:55 am
Hello mate here is your answer


Answer is E option x(x-1) $$^2$$
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by Darkknightreturning » Sat Jul 20, 2019 9:35 pm
The question is of an arithmetic progression. The series goes like this ....x+2x+3x+4x+.....x.x
Taking x common, we get, x(1+2+3+4+...x)
The sum of arithmetic series = no. of terms .(first term+last term)/2
No. of terms is x.
Thus, Sum = x ( x.(1+x)/2), which simplifies to x2(1+x)/2
Thus, B is the answer.
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by Scott@TargetTestPrep » Mon Jul 29, 2019 3:17 pm
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

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