Prove that for any natural n the value of the expression (4n+5)^2–9 is divisible by 4.
squaring (4 n + 5) ... 16 n^2 + 40 n + 25
subtracting 9 ... 16 n^2 + 40 n + 16
all the terms are divisible by 4
Oh, alright, hold on, let me grab my mathematician's jester hat for this one. 🤡
To prove that (4n+5)^2 - 9 is divisible by 4 for any natural number n, we need to demonstrate that the expression gives a remainder of 0 when divided by 4.
Let's begin:
(4n + 5)^2 - 9
= 16n^2 + 40n + 25 - 9
= 16n^2 + 40n + 16
= 4(4n^2 + 10n + 4)
Now, let's separate the original expression into a multiple of 4 and some additional terms:
4(4n^2 + 10n + 4)
As we can see, we have a multiple of 4 times an expression that contains only whole numbers. Therefore, this expression is indeed divisible by 4 for any natural number n.
And presto! (4n+5)^2 - 9 is proven to be divisible by 4. Hope I could put a smile on your face while proving some math! 🎉
To prove that the expression (4n+5)^2 - 9 is divisible by 4 for any natural number n, we need to show that the expression leaves a remainder of 0 when divided by 4.
Let's simplify the given expression step by step:
(4n+5)^2 - 9
Expanding the square:
= (16n^2 + 40n + 25) - 9
Combine like terms:
= 16n^2 + 40n - 9 + 25
= 16n^2 + 40n + 16
Factor out common factor 16:
= 16(n^2 + 2.5n + 1)
Notice that the expression inside the parentheses is n^2 + 2.5n + 1.
To prove that the expression is divisible by 4 for any natural number n, we need to show that n^2 + 2.5n + 1 is divisible by 4.
Let's consider the three possible cases:
Case 1: n is even
If n is even, it can be represented as 2k, where k is a natural number. Substituting this value into the expression:
(2k)^2 + 2.5(2k) + 1 = 4k^2 + 5k + 1
Since 4k^2, 5k, and 1 are all divisible by 4, the expression 4k^2 + 5k + 1 is also divisible by 4.
Case 2: n is odd, but not divisible by 4
If n is odd but not divisible by 4, it can be represented as 2k + 1, where k is a natural number. Substituting this value into the expression:
(2k + 1)^2 + 2.5(2k + 1) + 1 = 4k^2 + 8k + 4.5k + 4
The sum 4k^2 + 8k + 4.5k + 4 can be rearranged as (4k^2 + 8k + 4) + (0.5k), which is clearly divisible by 4 since 4k^2 + 8k + 4 is divisible by 4.
Case 3: n is divisible by 4
If n is divisible by 4, it can be represented as 4k, where k is a natural number. Substituting this value into the expression:
(4k)^2 + 2.5(4k) + 1 = 16k^2 + 10k + 1
Since 16k^2, 10k, and 1 are all divisible by 4, the expression 16k^2 + 10k + 1 is also divisible by 4.
Since the expression n^2 + 2.5n + 1 is divisible by 4 for all three cases (even, odd but not divisible by 4, and divisible by 4), we can conclude that (4n+5)^2 - 9 is divisible by 4 for any natural number n.
To prove that the expression (4n+5)^2 - 9 is divisible by 4 for any natural number n, we need to show that the expression can be written in the form 4a, where a is an integer.
Let's expand the given expression:
(4n + 5)^2 - 9
= (4n + 5)(4n + 5) - 9
= 16n^2 + 40n + 25 - 9
= 16n^2 + 40n + 16
Now, let's factor out 4 from the expression:
16n^2 + 40n + 16
= 4(4n^2 + 10n + 4)
As we can see, the expression 4n^2 + 10n + 4 is a multiple of 4 since every term contains a factor of 4.
Therefore, we have proven that (4n+5)^2 - 9 is divisible by 4 for any natural number n.
(4n+5)^2–9 = (4n+5)^2-3^2
= (4n+5+3)(4n+5-3)
= (4n+8)(4n-2)
= 4(n+2)(4n-2)
which is divisible by 4