The infinite sequence a1, a2, , an, is such that a_n = n! for n 0.

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 · Accepted 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

Aks111 · Answer

If we notice the answer options they all have different units digit.
With give information we have, a1 = 1! ; a2 = 2! = 2 ; a3 = 6 ; a4 = 24 ; a5 = 120 ; .......

Notice that, after a4 all numbers in the sequence will have '0' as the units digit.
Hence, the units digit of sum from a1 to a11 can be determined by adding units digit of numbers from a1 to a4 = 
1+2+6+4 = 13
Therefore, the units digit of the sum = 3, which matches D. 43954713

Answer: D

Ilanchezhiyan · Answer

Sir, how do we assume that n=11 in this question? It says sum of the first 11 terms; how do we know that the first term starts at 1?

Bunuel · Answer

We are given that the sequence starts with a1:

The infinite sequence  a_1, \ a_2, \ ..., \ a_n, \ ...

Ilanchezhiyan · Answer

Thanks sir, understood now.

uCS2020 · Answer

To solve this question, what I did:

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

Sequence = a1 , a2, a3, ... , aN ( N is any positive integer)

aN = N! = 1* 2 * 3* ... * N

Question asks for sum of first 11 terms of the sequence, meaning the sequence will look something like this:

1! , 2! , 3! , 4! , ... , 11!

And the sum of the terms is just 1! + 2! + 3! + ... + 11!

Now, finding the sum by just finding the answer to the factorial and adding it all is too much work.

Looking at the answer choices, all of the units digits are different and we can use this to our advantage to solve this question in almost an instant.

All the factorials from 5 onwards end in 0, but the factorials of 1 to 4 do not end in 0.

1! =1
2!=2 
3!=6 
4!=24 
Now just focusing on the units digits, 1 + 2 + 6 + 4 =13
And the units digits of 13 is 3
We already know that all the factorials from 5 onwards end in 0, meaning the units digit of the sum of these first 11 terms of the sequence is 3+0 = 3

Answer = D

ghimires28 · Answer

Bunuel 
My though process here, the final answer is odd as e+e+.....e+o=odd
so now between C. 43954675 D. 43954713 we need the number to be divisible by 3 as 3! and above all divisible and 1! +2! is 3 so we get answer choice D .
what you think with my approach.

vasu1104 · Answer

we need to find 1! + 2! + 3! + ..... + 10!+ 11!
from 5! onward each ends with 0
but some 1!+2!+3!+4! ends with 3.
so only option that has 3 in unit digit is D

Bunuel · Answer

Yes, that's correct.

The infinite sequence a1, a2, , an, is such that a_n = n! for n > 0.

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