prove that
2 2 2 2
(n-1) + n + (n+1) = 3n +2
Your statement
(n-1) + n + (n+1) = 3n +2
is false
(n-1) + n + (n+1)
= n-1+n+n+1
= 3n
I don't know what 2 2 2 2 is supposed to mean.
Looks like an attempt at exponents, not counting on html formatting:
(n-1)^2 + n^2 + (n+1)^2 = 3n^2 + 2
Now come on, guy. Just expand the expressions, and the +2n and -2n cancel out, and the +1 and +1 add up to 2.
Where do you have trouble here?
To prove the equation:
(n-1)^2 + n^2 + (n+1)^2 = 3n + 2
We will start by expanding each term and simplifying the equation step by step.
Expanding the terms:
(n-1)^2 = (n-1)(n-1) = n^2 - 2n + 1
(n+1)^2 = (n+1)(n+1) = n^2 + 2n + 1
Substituting the expanded forms:
(n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1) = 3n + 2
Now, combining like terms:
n^2 + n^2 + n^2 - 2n + 2n + 1 + 1 = 3n + 2
Rearranging the terms:
3n^2 + 2 = 3n + 2
Simplifying:
3n^2 + 2 - 3n - 2 = 0
Combining like terms:
3n^2 - 3n = 0
Finally, factoring out n:
3n(n - 1) = 0
Now, we have two solutions for n that make the equation true:
(1) n = 0
(2) n = 1
By substituting these values back into the equation, we can verify that both sides are equal, confirming that the original equation holds true.