What is the value of the expression below: 1*2*3 + 2*3*4 + 3*4*5 + 4*5

Question

What is the value of the expression below:

1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + 5*6*7 + 6*7*8 + 7*8*9 + 8*9*10 + 9*10*11 + 10*11*12 ?

A. 2,680
B. 3,680
C. 4,290
D. 5,720
E. 6,170

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lacktutor · Accepted Answer

6+ 24+ 60 + 120 + ....+ 1320 ( divisible by 6 )

All answer options are divisible by 2. Need to check whether they are divisible by 3 or not:

A. 2,680 (2+6+8+0 = 16)
B. 3,680 (3+6+8+0 = 17)
C. 4,290  (4+2+9+0 = 15) 
D. 5,720 (5+7+2+0 = 14)
E. 6,170 (6+1+7+0 = 14)

Only answer option C is divisible by 6 
Answer (C)

preetamsaha · Answer

1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + 5*6*7 + 6*7*8 + 7*8*9 + 8*9*10 + 9*10*11 + 10*11*12 ?

Tn= n(n+1)(n+2) = (n^2+n)(n+2) = (n^3+3n^2+2n)
Sn= Σ(n^3+3n^2+2n) 
=Σ(n^3) + Σ(3n^2) + Σ(2n)
=Σ(n^3) + 3Σ(n^2) + 2Σ(n)
= ^2 + 3*  + 2* 
=55^2 + 3*11*35 + 2*11*5
=55* 
=55*78
=4290
 
correct answer C

note: Sum of First n Natural Numbers = 1+ 2+ ... + n = n(n+1) / 2
Sum of squares of First n Natural Numbers = 1^2+ 2^2+ ... + n^2 = n(n+1)(2n+1) / 2
Sum of cubes of First n Natural Numbers = 1^3+ 2^3+ ... + n^3 =  ^2

SarinaM · Answer

Sum of squares of First n Natural Numbers = 1^2+ 2^2+ ... + n^2 = n(n+1)(2n+1) / 2
I think you made a typo. It should be Sum of squares of First n Natural Numbers = 1^2+ 2^2+ ... + n^2 = n(n+1)(2n+1) / 6 instead.

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Archit3110 · Answer

for given expression
1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + 5*6*7 + 6*7*8 + 7*8*9 + 8*9*10 + 9*10*11 + 10*11*12
least common value is 6
so we can re write it as
6*( 1+4+10+20+35+56+84+120+165+220)
6*(175) 
= 4290
OPTION C

ScottTargetTestPrep · Answer

Solution:

Since the answer choices all have a different thousands digit, they are not close to one another. So we can just estimate the given sum. Furthermore, notice that each term in the given sum is approximately the middle factor raised to the third power (1*2*3 ≈ 2^3, 2*3*4 ≈ 3^3, etc.). Therefore, the given sum is approximately:

2^3 + 3^3 + … + 11^3

Let’s also throw in 1^3 at the beginning of this approximate sum:

1^3 + 2^3 + 3^3 + … + 11^3

We now can use the following formula:

1^3 + 2^3 + 3^3 + … + n^3 =  ^2

So, 1^3 + 2^3 + 3^3 + … + 11^3 =  ^2 = 66^2 = 4,356.

We see that the correct answer must be 4,290 since it’s the closest to the estimated sum of 4,356.

Alternate Solution:

If you aren’t familiar with the special formula given in the previous solution, we can still get the correct answer by estimating. Our strategy is to pull out the common factor from each successive pair of terms and estimate the sum of each pair.

Factor out 6 from the first two terms 1*2*3 + 2*3*4 to get 6(1 + 4) = 6 * 5 = 30

Factor out 20 from 3*4*5 + 4*5*6 to get 20(3 + 6) = 20 * 9 = 180 ≈ 200

Factor out 42 from 5*6*7 + 6*7*8 to get 42(5 + 8) = 42 * 13 ≈ 500

Factor out 72 from 7*8*9 + 8*9*10 to get 72(7 + 10) = 72 * 17 ≈ 1400

Factor out 110 from 9*10*11 + 10*11*12 to gt 110(9 + 12) ≈ 2300

The estimated and rounded subtotals can easily be added to obtain 4,430, which is closest to 4,290.

Answer: C

IanStewart · Answer

Each term in the sum is the product of three consecutive integers. Among any three consecutive integers, you always have one multiple of 3, so each term in the sum is a multiple of 3. When we add multiples of 3, we must get a multiple of 3. So the right answer must be divisible by 3. Summing the digits of each answer choice, only C is divisible by 3, so it must be correct.

Kinshook · Answer

Asked: What is the value of the expression below:

1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + 5*6*7 + 6*7*8 + 7*8*9 + 8*9*10 + 9*10*11 + 10*11*12 ?

t_n = n(n+1)(n+2) =\frac{ ((n+3)-(n-1))*n(n+1)(n+2)}{4} = \frac{n(n+1)(n+2)(n+3)}{4} - \frac{(n-1)n(n+1)(n+2)}{4}

1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + 5*6*7 + 6*7*8 + 7*8*9 + 8*9*10 + 9*10*11 + 10*11*12 =  \frac{10*11*12*13}{4} - 0 = 4290 
Since other terms cancel out

IMO C

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What is the value of the expression below: 123 + 234 + 345 + 4*5

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What is the value of the expression below: 123 + 234 + 345 + 4*5