M39-02

Question

The infinite sequence  a_1, \ a_2, \ …, \ a_n, \ …  is such that  a_n = n!  for  n > 0 . What is the sum of the first 11 terms of the sequence?

A.  43954414 
B.  43954588 
C.  43954675 
D.  43954713 
E.  43954780

Bunuel · Answer

Official Solution:

The infinite sequence  a_1, \ a_2, \ …, \ a_n, \ …  is such that  a_n = n!  for  n > 0 . What is the sum of the first 11 terms of the sequence?

A.  43954414 
B.  43954588 
C.  43954675 
D.  43954713 
E.  43954780

According to the sequence formula given, the sequence is:  1!, \ 2!, \ 3!, \ ...  . We are asked to find the value of:
  
 1!+2!+3!+4!+5!+6!+7!+8!+9!+10!+11! .
  
Direct calculation will take long time and will be quite tiresome. So, what to? 
 
First notice that the  units digits  of answer choices are different. Next, notice that from 5! onwards the units digit of all terms will be 0 (because each of them will have a 2 and a 5 in it). So, to get the units digit of the sum we only need to calculate the units digit of  1!+2!+3!+4! , which is 3 ( 1!+2!+3!+4!=1+2+6+24=33 ). 
 
Thus, the units digit of  1!+2!+3!+4!+5!+6!+7!+8!+9!+10!+11!  will also be 3. The only option which has a number with the units digit of 3 is D.

Answer: D

chetan2u · Answer

Just for the sake of giving another way, here is one more, although the method of using units digit as shown by  Bunuel  will help you almost always.

1!+2!+3!+……….11!
 First observation:  All terms starting from 2! will be even, so adding 1 to the sum will make it ODD. Only C and D remain. 
 Second observation:  All terms starting from 3! will be multiple of 3, so adding 1!+2! or 3 to the sum will keep it multiple of 3.
C. 4+3+9+5+4+6+7+5=43. Not a multiple of 3, so discard. 
D. 4+3+9+5+4+7+1+3=36. We have a multiple of 3, so that’s our answer.

D

nikitathegreat · Answer

How do we get to know that they are forming the sequence of 1!, 2!, 3! etc. If I take n=3 then I know that a3= 3!. How do I know what will be the value of a1, a2? Is it because is says sequence and hence a1 and a2 will be 1! and 2!?
Thanks

Bunuel · Answer

We are given that  a_n = n! , so by substituting  n  for the number indicating the term’s place in the sequence, you get its value:

a_1 = 1! 
 a_2 = 2! 
 a_3 = 3!

and so on.

It's that simple!

nikitathegreat · Answer

I somehow thought an = last term and not putting value of n =1,2 and getting factorial as output. So, I thought we only know the last value of the sequence. Can you explain this? Also, I see dots after an now, this means this cannot be the last term? Thanks

Bunuel · Answer

a_n  represents the nth term of the sequence, which is the standard way of referring to a term's position in a sequence. And yes, the dots after  a_n  signify that the sequence continues indefinitely. Additionally, since it is explicitly mentioned that the sequence is infinite, it does not have a last term—it goes on forever!

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 Sequences Made Easy - All in One Topic! 
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For other subjects:
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 Ultimate GMAT Quantitative Megathread

Hope it helps.

sunshineeee · Answer

First, get the question right.
Sum of first 11 term of sequence means: 1! + 2! + 3! + ... + 11! = ?

Second, try to get the pattern -> see the answers option has different last unit digits, focus on this

1! = 1
2! = 2
3! = 3*2*1 = 6
4! = 4*3*2*1 = 4*6 = 24
5! = 5*4! = 5*24 = 120
if it get total --> 1+2+6+24=...3+120=...3 (3 as the last unit digit)
So till 5! above will have trailling zeros, so, final answer will be end with 3.

Thus, answer is  D. 43954713

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