Prove that for any natural value of n the value of the expression:
(5n+1)2–(2n–1)2 is divisible by 7
3n^2+2n * 7
thats the answer
assume you mean
(5n+1)^2–(2n–1)^2
25 n^2 + 10 n + 1 - [ 4 n^2 -4 n + 1 ]
= 21 n^2 + 14 n + 0
= (1/7)(3 n^2 + 2 n)
To prove that the expression (5n+1)2 - (2n-1)2 is divisible by 7 for any natural value of n, we can simplify the expression and show that it is divisible by 7.
Let's begin by expanding the two squares:
(5n + 1)^2 - (2n - 1)^2
= (25n^2 + 10n + 1) - (4n^2 - 4n + 1)
= 25n^2 + 10n + 1 - 4n^2 + 4n - 1
= 21n^2 + 14n
Now, we need to show that this expression is divisible by 7, which means there must exist an integer k such that 21n^2 + 14n = 7k.
Factoring out 7 from the expression:
21n^2 + 14n
= 7(3n^2 + 2n)
Therefore, we have shown that the expression (5n + 1)^2 - (2n - 1)^2 is divisible by 7 for any natural value of n, since we can rewrite it as 7 multiplied by an integer (3n^2 + 2n).
To prove that for any natural value of n, the expression (5n+1)^2 – (2n–1)^2 is divisible by 7, we will use the method of mathematical induction.
Step 1: Base Case
Let's check if the expression is divisible by 7 for the lowest possible value of n, which is 1.
Substituting n = 1 into the expression, we have:
(5(1) + 1)^2 – (2(1) – 1)^2
= (6)^2 – (1)^2
= 36 – 1
= 35
Since 35 is divisible by 7 (7 * 5 = 35), the base case holds.
Step 2: Inductive Hypothesis
Assume that the expression is divisible by 7 for an arbitrary natural number k, denoted as P(k).
Step 3: Inductive Step
We need to prove that P(k) ➝ P(k+1), which means if the expression is divisible by 7 for k, it is also divisible by 7 for k+1.
Substituting n = k+1 into the given expression, we have:
(5(k+1) + 1)^2 – (2(k+1) – 1)^2
= (5k + 5 + 1)^2 – (2k + 2 – 1)^2
= (5k + 6)^2 – (2k + 1)^2
= 25k^2 + 60k + 36 – 4k^2 – 4k + 1
= 21k^2 + 56k + 37
Now, we want to show that 21k^2 + 56k + 37 is divisible by 7.
Using the assumption P(k), we know that 21k^2 + 49k is divisible by 7.
Therefore, we can rewrite 21k^2 + 56k + 37 as (21k^2 + 49k) + (7k + 37).
Notice that 7k + 37 is a multiple of 7 since both 7k and 37 are divisible by 7.
Therefore, (21k^2 + 56k + 37) is divisible by 7.
Step 4: Conclusion
By the principle of mathematical induction, we have shown that for any natural value of n, the expression (5n+1)^2 – (2n–1)^2 is divisible by 7.