# Prove that the number a is natural. a = (2^(2n+3)*3^(n+2)*5^(2n+1) + 3^(n+3)*4^(n+2)*5^(2n))/792 n is from N set

Tushar Chandra | Certified Educator

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We have a = (2^(2n+3)*3^(n+2)*5^(2n+1) + 3^(n+3)*4^(n+2)*5^(2n))/792

=> (2^3*2^2n*3^2*3^n*5^1*5^2n + 3^n*3^3*4^n*4^2*5^2n)/792

=> (8*9*5*2^2n*3^n*5^2n + 27*16*3^n*4^n*5^2n)/792

=> 2^2n*3^n*5^2n (8*9*5 + 27*16* )/ 792

=> 2^2n*3^n*5^2n*792/ 792

=> 2^2n*3^n*5^2n

Therefore a = 2^2n*3^n*5^2n

Now as n is from a natural set , we can say that a is natural.

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## Related Questions

neela | Student

To prove that the number a is natural.

a = (2^(2n+3)*3^(n+2)*5^(2n+1) + 3^(n+3)*4^(n+2)*5^(2n))/792.

a = 2^(2n+3)*3^(n+2)+3^(n+3)*2^(2n+4)*5^2n)/732, where we replaced 4^(n+2)  by (2^2)(n+2) = 2^(2n+4).

a = 2^(2n+3)* 3^(n+2)*5^2n{5 + 3*2*5^2n}/792

a = 2^3*3^2 * (2^2*3*5^2)^n{5+6*25^n}792

a = 300^n{5+6}/11

a = 300^n  which belongs to N.

giorgiana1976 | Student

We notice that we can find common factors at numerator. We'll factorize by the numbers that has the lowest powers.

First, we'll compare the powers:

3^(n+2) < 3^(n+3)

5^(2n + 1)<5^(2n)

For the beginning, we'll factorize by 3^(n+2)*5^(2n)

a = 3^(n+2)*5^(2n)*[2^(2n + 3)*5 + 3*4^(n+2)]/792

We notice that we can write 4^(n+2) as a power of 2:

4^(n+2) = 2^2(n + 2) = 2^(2n + 4)

2^(2n + 4) > 2^(2n + 3)

We'll factorize by 2^(2n + 3):

a = 2^(2n + 3)*3^(n+2)*5^(2n)*(5 + 6)/792

a = 2^(2n + 3)*3^(n+2)*5^(2n)*(11)/792

We'll simplify by 11:

a = 2^(2n + 3)*3^(n+2)*5^(2n)/72

We'll re-write the numerator:

2^(2n + 3) = 2^2n*2^3 = 8*2^2n

3^(n+2) = 3^n*3^2 = 9*3^n

2^(2n + 3)*3^(n+2) = 8*9*2^2n*3^n

2^(2n + 3)*3^(n+2) = 72*2^2n*3^n

a = 72*2^2n*3^n*5^2n/72

We'll simplify by 72:

a = 2^2n*3^n*5^2n

Since n is natural, then 2^2n is natural, too.

Since n is natural, then 3^n is natural, too.

Since n is natural, then 5^2n is natural, too.

The product of 3 natural numbers is also a natural number:

a = 2^2n*3^n*5^2n is a natural number.