If x is a positive integer greater than 1, what is the sum of the mult

Question

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

ENGRTOMBA2018 · Accepted Answer

Algebraically:

You need to find the sum : x+2x+3x+4x+5x+6x......x*(x-1)+x*x = x(1+2+3+4+5+6....+x-1+x) = x* (sum of numbers from 1 to x) =  x* (x/2)    .......(from sum of n terms of an arithmetic progression :  S_n  =  \frac{n }{2}  , where n = total number of terms, a = 1st term, d = difference between 2 consecutive terms

= \frac{x^2(x+1)}{2}

B is the correct answer.

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

We can approach this problem with algebra or by plugging numbers. Even though the latter’s probably faster and easier, it’s good to have both tools up your sleeve.

Algebraic solution (harder):

The multiples of x are x, 2x, 3x, etc. The square of x is also a multiple of x, of course—it’s just x times x, or the xth multiple of x.

So the list we care about is x, 2x, 3x… up to x^2.

The sum of these numbers can be written as x + 2x + 3x + … + x^2.

We can now factor out an x to get x(1 + 2 + 3 + … + x).

Now, the sum that remains (that is, 1 + 2 + 3 + … + x) is a sum of consecutive integers, which is evenly spaced.

It’s easy to calculate the average (arithmetic mean) of an evenly spaced set: just add up the outermost numbers and divide by 2: Average of {1, 2, 3, …, x} = (x + 1)/2

The number of numbers in that set is just x, since there are x consecutive integers between 1 and x, inclusive. (That sounds harder than it is! If x is 3, then all we’re saying is that in the set {1, 2, 3}, there are 3 numbers.)

Now, back to {1, 2, 3, …, x}. Since the average equals the sum divided by the number of numbers, the sum equals the average times that number of numbers.

So 1 + 2 + 3 + … + x = Average × Number =  x

Finally, to get the original sum, x + 2x + 3x + … + x^2, we just multiply by x again to get x^2(x + 1)/2.

Plugging numbers:

Pick x = 2. The sum of multiples of 2 from 2 to 2^2 is just 2 + 4 = 6. Check the answers:

(A) 2(3)(1) = 6
(B) 2^2(3 + 1)/2 = 6
(C) 2^2(2-1)=4
(D) (2^3 + 2*2)/2 = 6
(E) 2(2-1)^2=2

Okay, all we can eliminate is C and E. Try x = 3. The sum of multiples of 3 from 3 to 3^2 is just 3 + 6 + 9 = 18. Check the remaining answers:

(A) 3(4)(2) = 24
(B) 3^2(3 + 1)/2 = 18
(D) (3^3 + 2*3)/2 = 16.5

goldfinchmonster · Answer

Ans :- B.
I substituted 3 in all answer choices.
Option B & C were both satisfying the condition.

So i substituted 4 instead of 3 in all answer choices,
Only B option satisfied.

Can anyone tell a simpler way to solve.

kunal555 · Answer

We have to find the sum of multiples of x from x to x^2 inclusive

let x = 3
then sum of the multiples from 3 to 9 will be 
3+6+9
or 3(1+2+3)

let x=5
then sum of multiples from 5 to 25 will be
5+10+15+20+25
or 5(1+2+3+4+5)

From these values it is clear that th sum of multiples of x from x to x^2 inclusive will be
x(sum of first x natural numbers)
or  x*   \frac{x(x+1)}{2} 
or  \frac{x^2(x+1)}{2}

Answer:- B

anudeep133 · Answer

Solution:

Sum = x + 2x +3x +.. + x(x) = x(1+2+3...+x) = x((x/2)(2 +(x-1)) = (x^2)(x+1)/2
So, option B

GMATinsight · Answer

Let, x=2

sum of the multiples of x from x (i.e. 2) to x^2 (i.e. 2^2=4), inclusive = 2+4 = 6

@x=2
 (A) x(x + 1)(x-1) = 2*3*1 = 6
(B) x^2(x + 1)/2 = 2^2*3/2 = 6 
 (C) x^2(x-1) = 2^2*3 = 12 
 (D) (x^3 + 2x)/2 = (8+4)/2 = 6 
 (E) x(x-1)^2 = 2*1 = 2

i.e. Possible Options are only A, B and D

Now, Let, x=3

Sum of the multiples of x from x (i.e. 3) to x^2 (i.e. 3^2=9), inclusive = 3+6+9 =  18

@x=3
(A) x(x + 1)(x-1) = 3*4*2 = 24
 (B) x^2(x + 1)/2 = 3^2*4/2 = 18 
(D) (x^3 + 2x)/2 = (27+6)/2 = 33/2

i.e. Answer Option B

stonecold · Answer

Here is what i did in this one =>Putting x=5
Sum = 5+10+15+20+25=75
only option that matches is B.

Hence B.

ScottTargetTestPrep · Answer

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 between the 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

Answer: B

SchruteDwight · Answer

This question is about series.

Either through trial and error, or through

a_n=a_1+(n-1)d \implies x^2=x+(n-1)x

you'll find that  n=x

therefore, since we are dealing with an equally spaced set,

average*n=sum \implies \frac{(x+x^2)}{2}*x= \frac{(1+x)x^2}{2}

bumpbot · Answer

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If x is a positive integer greater than 1, what is the sum of the mult

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