Question

prove a(a+1)(a+2)(a+3)=2k(2k+2)

Answer 1

Equation given :

a(a+1)(a+2)(a+3) = 2k(2k+2)

a²+a(a+2)(a+3) = 2k(2k+2)

a³+2a²+a²+2a(a+3) = 2k(2k+2)

a³+3a²+2a(a+3) = 4k²+4k

a⁴+3a³+2a²+3a³+9a²+6a = 4k²+4k

a⁴+6a³+11a²+6a = 4k²+4k

As you can't see that no proper solution is found, so there must be some other data also given, which could have been missed out, so please check the question again, because this question is incomplete.

Answer 2

To Prove :

Prove that the product of any four consecutive integers can be written as the product of two consecutive even numbers .

Proof :

We need to take two cases :

(a) , the starting integer of the four consecutive integers are odd

(b) the starting integer of the  four consecutive integers are even

[a] Let the integers be :

> 2k-1 , 2k, 2k+1 and 2k+2

Product :

> (2k-1)2k(2k+1)(2k+2)

> 2k * ( 4k^2 - 1)( 2k + 2)

> 4k(4k^2 - 1)( k + 1)

> [ 2k ( 4k^2 - 1) ] * [ 2 ( k + 1)]

This can be written as a product of two consecutive even integers,

[b] Let the integers be :

> 2k-2, 2k-1, 2k and 2k+1

Product :

> (2k-2)(2k-1)2k(2k+1)

> 2(k-1)2k(4k^2 -1 )

> [ 2k( 4k^2 - 1) ] * [ 2(k-1) ]

This can be written as a product of two consecutive even integers .

Hence Proved

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