What is the value of (n+1)!?

Question

I have seen the below value in the text book. Can someone explain this equation?
(n+1)! = (n+1)(n)(n-1)

Answer 1

　

x! = x * (x-1) * (x-2) * ... * 3 * 2 * 1



But since multiplying by 1 changes nothing, we could also say that:



x! = x * (x-1) * (x-2) * ... * 3 * 2



Therefore, since (n+1)! = (n+1) * n * (n-1)

then (n-1) = 1 or 2



n = 2 or 3

Answer 2

n+1)!=(n+1)*n*(n-1)*(n-2)*.........................3*2*1

Answer 3

Depends on the value you assign to 'n', usually a variable. It helps calculation with variables easy.

Answer 4

For example 3!= 3x2x1= 3x(3-1)x(3-2)

So, (n+1)= (n+1)x(n+1-1)x(n+1-2)

= (n+1)x(n)x(n-1)

Answer 5

n=3

Answer 6

(n + 1)! = (n + 1)n(n − 1)

given (n + 1)! = (n + 1)n!

∴ (n + 1)n! = (n + 1)n(n − 1)

given n! = n(n − 1)!

∴ (n + 1)n(n − 1)! = (n + 1)n(n − 1)

given (n − 1)! = (n − 1)(n − 2)!

∴ (n + 1)n(n − 1)(n − 2)! = (n + 1)n(n − 1)

∴ (n + 1)n(n − 1)(n − 2)! − (n + 1)n(n − 1) = 0

∴ (n + 1)n(n − 1)[(n − 2)! − 1] = 0

∴ n + 1 = 0, n = 0, n − 1 = 0 and (n − 2)! − 1 = 0

∴ n = -1, n = 0, n = 1 and (n − 2)! = 1

given 1! = 1 and 0! = 1

∴ n = -1, n = 0, n = 1, n − 2 = 1 and n − 2 = 0

∴ n = -1, n = 0, n = 1, n = 2 and n = 3

now factorials are only defined whole numbers not integers so in the original form of the equation tells n + 1 ≥ 0 and (n + 1)n(n − 1) ≥ 1

as n=-1,0 and 1 all make (n + 1)n(n − 1) ≥ 1 false they are not valid solutions

∴ n = 2 and n = 3



test

n = 3 then (n + 1)! = (n + 1)n(n − 1) is 4! = 4*3*2 is true.

n = 2 then (n + 1)! = (n + 1)n(n − 1) is 3! = 3*2*1 is true.

n = 1 then (n + 1)! = (n + 1)n(n − 1) is 2! = 2*1*0 is false.

n = 0 then (n + 1)! = (n + 1)n(n − 1) is 1! = 1*0*-1 is false.

n = -1 then (n + 1)! = (n + 1)n(n − 1) is 0! = 0*-1*-2 is false.

Answer 7

this equation: (n+1)! = (n+1)(n)(n-1),So,

By Inspection, n = 3,

So,

(3+1)! = (3+1)(3)(3-1), == > 4! = 4*3*2,

Or,

4*3*2*1 = 24, ==> 24 = 24,

Hence,

n = 3. <------------------------------/

Answer 8

22

Answer 9

Not quite.

(n+1)! = (n+1)(n)(n-1)!



(n+1)! = (n+1)(n)(n-1)(n-2)(n-3)....(3)(2)(1)

But (n-1)(n-2)(n-3)....(3)(2)(1) = (n-1)!

Answer 10

n = 3

Answer 11

n+1)!=(n+1)*n*(n-1)*(n-2)*...................................................

Answer 12

when a rule,law,etc apply to any number ,the number is represented by the symbol " n "......... it is different that using " x " in algebra for the reason that x represents a exact number, while n represents one value at a time that follows a rule......... it is used when dealing with series or sequences

for example......... " write the sum of all the whole numbers from 1 to 5 "......... considering that mathematics is a " universal language" we use symbols to replace the written language like english.........

the greek letter sigma that looks like m but sidewise ( a big s ) represents summation ( the sum of ........................... )......... the example above is written :

................................................................................................................................. = 5

................................................................................................................................. the symbol sigma = 1 + 2 + 3 + 4 + 5 ............................................. we could call 1 the lower limit and 5 the upper.........limit

................................................................................................................................. n= 1



what about ( n + 1 ) ! this shall include the next number after the upper lime.........the lower limit also increases by 1



................................................................................................................................. n = 5

for example : ........................... sigma ( n +1 ) = 2 + 3 + 4 + 5 + 6

................................................................................................................................. n = 1



i hope this helps



when we write " n factorial " we are using n in the same sense......... 4! = 1 x 2 x 3 x 4

Answer 13

(n+1)! = (n+1)(n)(n-1)...[1]. Note: This is an equation, so we are required to find the

value of n such that this equation holds.

In general, by definition, (n+1)! = (n+1)(n)(n-1)(n-2)(n-3).......(6)(5)(4)(3)(2)(1). The last digit of the

factorial function, by definition, is 1. Therefore, (n-1) = 1, ie., n = 2 & [1] becomes 3! = 3*2*1. If one accepts n = 3 as a solution, then [1] becomes 4! = 4*3*2 which is technically incorrect even though 4*3*2 = 4*3*2*1.

Answer 14

6 gorillion

Answer 15

22

Answer 16

n is a symbol & not a number. Unless you attribute (give) some value to it you can't ask the 'value' of (n+1), you idiot !

Your expression (n+1)! =(n+1)(n)(n-1) is fcuking wrong. A correct expression is

(n+1)! =(n+1)(n)(n-1)(n-2)(n-3)(n-4) . . . . . (4)(3)(2)(1).

See that the expression of muliplicants ends in 1.That is the meaning of (n+1)!

Answer 17

When a rule,law,etc apply to any number ,the number is represented by the symbol " n ". It is different that using " x " in Algebra for the reason that x represents a exact number, while n represents one value at a time that follows a rule. It is used when dealing with series or sequences

For example. " Write the sum of all the whole numbers from 1 to 5 ". Considering that Mathematics is a " Universal Language" we use symbols to replace the written language like English.

The greek letter sigma That looks like M but sidewise ( a big S ) represents Summation ( the sum of ... ). The example above is written :

............................................n = 5

........................................ The symbol Sigma = 1 + 2 + 3 + 4 + 5 ..... We could call 1 the lower limit and 5 the upper.limit

........................................... n= 1



What about ( n + 1 ) ! this will include the next number after the upper lime.The lower limit also increases by 1



......................... n = 5

For example : ... SIGMA ( n +1 ) = 2 + 3 + 4 + 5 + 6

......................... n = 1



I hope this helps



When we write " n factorial " we are using n in the same sense. 4! = 1 x 2 x 3 x 4

Answer 18

n=3

Answer 19

(n+1)! = (n+1)n(n-1)...*3*2*1, where n equals 0 or is a whole positive number.

Answer 20

It's one more than you had before, so if n = 3 apples, n+1 = 4 apples, OK?

Answer 21

One more!

Answer 22

Lhs#0--->rhs#0

--->n>=2

(i) n=2 is ok

(ii) n>2 divide both sides by (n-1)n(n+1)

--->(n-2)!=1

---> n=3

solutions : 2,3

Answer 23

(n + 1)! = (n + 1) * n * (n - 1) is not an identity.



However, if you want it worked out:



(n + 1)! =>

(n + 1) * n * (n - 1) * (n - 2) * .... * 3 * 2 * 1



If (n + 1)! = (n + 1) * n * (n - 1), then:



(n + 1) * n * (n - 1) * (n - 2)! = (n + 1) * n * (n - 1)

(n - 2)! = 1

n - 2 = 0 , 1

n = 2 , 3



Test:



(3 + 1)! = 4! = 24

(3 + 1) * 3 * (3 - 1) = 4 * 3 * 2 = 24



(2 + 1)! = 3! = 6

(2 + 1) * 2 * (2 - 1) = 3 * 2 * 1 = 6

Answer 24

(n+1)! = (n+1)(n)(n-1).....(2)(1)

Like all factorials.