Check the forward differences of the sequence.
If
, then let
be the sequence of first-order differences of
. That is, for n ≥ 1,

so that
.
Let
be the sequence of differences of
,

and we see that this is a constant sequence,
. In other words,
is an arithmetic sequence with common difference between terms of 2. That is,

and we can solve for
in terms of
:



and so on down to

We solve for
in the same way.

Then



and so on down to


X=4 this is because 4 times (x+1)-5=11 the 1x then just turns to x because you can’t have a 1 next to x when it’s just one x.
4*(x+1)-5=11
4(x)-5=11
4x-5=11
+5=11
—————
4x=16
4. 4
———-
X=4
If you have choices, you should list them. If you are just asking about any example, then here are some.
y = a*x + 3 ; the value of a can be any real number.
- 3x + y = 3
2y + 4x = 6
ny = ax + 3*n When you divide through by n, you get
y = a/n * x + 3n/n
y = ax/n + 3
Expression equivalent to
is ![(\sqrt[d]{3b})^2](https://tex.z-dn.net/?f=%28%5Csqrt%5Bd%5D%7B3b%7D%29%5E2)
Option D is correct.
Step-by-step explanation:
We need to find equivalent expression of: 
Solving:
We know that ![\frac{1}{d}= \sqrt[d]{}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bd%7D%3D%20%5Csqrt%5Bd%5D%7B%7D)
So, the expression will become:

![=(\sqrt[d]{3b})^2](https://tex.z-dn.net/?f=%3D%28%5Csqrt%5Bd%5D%7B3b%7D%29%5E2)
So, expression equivalent to
is ![(\sqrt[d]{3b})^2](https://tex.z-dn.net/?f=%28%5Csqrt%5Bd%5D%7B3b%7D%29%5E2)
Option D is correct.
Keywords: Exponents
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