Question

what's next number this series​

Answer 1

120

1st term in series: 1

2nd term = 1st term (1)=1

3rd term = 2nd term * 2= 2

4th = 3rd term*3=6

5th = 4th term*4 = 24

6th = 5th term*5 = 120

Answer 2

This is the sequence of each factorial value of the whole numbers!

The nth term of this sequence would be (n - 1)!

But what is (n - 1)! ?

For any positive integer k,

$\displaystyle \Large k! = 1 \times 2 \times 3 \times 4 \times ...... \times k \\ \\ \\ k! = \prod_{i=1}^{k}i$

Thus, taking k = n - 1,

$\displaystyle \Large (n-1)!=1 \times 2 \times 3 \times 4 \times ...... \times (n-1) \\ \\ \\ (n-1)!\ =\ \prod_{i=1}^{n-1}i\ =\ \prod_{i=2}^{n}(i-1)$

There's no negative values for k. The value of 0! is 1.

Now, consider the given sequence.

$\displaystyle \large T_1=(1-1)!=0!=\ \Large \text{1} \\ \\ \\ \large T_2=(2-1)!=1!=\ \Large \text{1} \\ \\ \\ \large T_3=(3-1)!=2!=\ \Large \text{2} \\ \\ \\ \large T_4=(4-1)!=3!=\ \Large \text{6} \\ \\ \\ \large T_5=(5-1)!=4!=\ \Large \text{24}$

Thus the next term, 6th term will be,

$\displaystyle \large T_6=(6-1)!=5!=\ \Large \textbf{120}$

Hence 120 is the answer.

Each consecutive terms in the sequence are multiplied up by the natural numbers in order, like,

$\large \text{1} \times 1 = \large \text{1} \\ \\ \\ \large \text{1} \times 2 = \large \text{2} \\ \\ \\ \large \text{2} \times 3 = \large \text{6} \\ \\ \\ \large \text{6} \times 4 = \large \text{24} \\ \\ \\ \large \text{24} \times 5 = \large \textbf{120}$

$\displaystyle \large \textsf{What about writing the \ n^{\textsf{th}} term as \ ^{n-1}\!P_{n-1} \ ?!} \\ \\ \\ \large T_1=\ ^{1-1}\!P_{1-1} =\ ^0\!P_0 = \frac{0!}{(0-0)!}=\frac{0!}{0!}=\frac{1}{1}=\ \Large 1 \\ \\ \\ \large T_2=\ ^{2-1}\!P_{2-1} =\ ^1\!P_1 = \frac{1!}{(1-1)!}=\frac{1!}{0!}=\frac{1}{1}=\ \Large 1 \\ \\ \\ \large T_3=\ ^{3-1}\!P_{3-1} =\ ^2\!P_2 = \frac{2!}{(2-2)!}=\frac{2!}{0!}=\frac{2}{1}=\ \Large 2$

$\displaystyle \large T_4=\ ^{4-1}\!P_{4-1} =\ ^3\!P_3 = \frac{3!}{(3-3)!}=\frac{3!}{0!}=\frac{6}{1}=\ \Large 6 \\ \\ \\ \large T_5=\ ^{5-1}\!P_{5-1} =\ ^4\!P_4 = \frac{4!}{(4-4)!}=\frac{4!}{0!}=\frac{24}{1}=\ \Large 24 \\ \\ \\ \\ \large \textsf{Thus,}\\ \\ \\ \large T_6=\ ^{6-1}\!P_{6-1} =\ ^5\!P_5 = \frac{5!}{(5-5)!}=\frac{5!}{0!}=\frac{120}{1}=\ \Large \textbf{120}$