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04 Nov 2022, 08:50
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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. \(43954780\)
A. \(43954414\)
B. \(43954588\)
C. \(43954675\)
D. \(43954713\)
E. \(43954780\)
04 Nov 2022, 08:50
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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
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
04 Nov 2022, 09:05
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Bunuel
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
