Explicit Formula | geometric sequence formula

Explicit Formula-How explicit formulas work

Explicit Formula,An explicit formula designates the nth term of the sequence, as an expression of n (where n = the term’s location). It defines the sequence as a formula in terms of n

Here is an explicit formula of the sequence 3, comma, 5, comma, 7, comma, point, point, point
a, left parenthesis, n, right parenthesis, equals, 3, plus, 2, left parenthesis, n, minus, 1, right parenthesis
In the formula, n is any term number and a, left parenthesis, n, right parenthesis is the n, start superscript, start text, t, h, end text, end superscript term.
This formula allows us to simply plug in the number of the term we are interested in, and we will get the value of that term.
In order to find the fifth term, for example, we need to plug n, equals, 5 into the explicit formula.
\begin{aligned}a(\greenE 5)&=3+2(\greenE 5-1)\\\\ &=3+2\cdot4\\\\ &=3+8\\\\ &=11\end{aligned}
Cool! This is in fact the fifth term of 

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Equivalent explicit formulas

Explicit formulas can come in many forms.
For example, the following are all explicit formulas for the sequence 3, comma, 5, comma, 7, comma, point, point, point
  • 3, plus, 2, left parenthesis, n, minus, 1, right parenthesis (this is the standard formula)
  • 1, plus, 2, n
  • 5, plus, 2, left parenthesis, n, minus, 2, right parenthesis
The formulas may look different, but the important thing is that we can plug an n-value and get the correct n, start superscript, start text, t, h, end text, end superscript term (try for yourselves that the other formulas are correct!).
Different explicit formulas that describe the same sequence are called equivalent formulas.

Explicit Formula A common misconception

An arithmetic sequence may have different equivalent formulas, but it’s important to remember that only the standard form gives us the first term and the common difference.
For example, the sequence 2, comma, 8, comma, 14, comma, point, point, point has a first term of start color #0d923f, 2, end color #0d923f and a common difference of start color #ed5fa6, 6, end color #ed5fa6.

The explicit formula start color #0d923f, 2, end color #0d923f, start color #ed5fa6, plus, 6, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis describes this sequence, but the explicit formula start color #0d923f, 2, end color #0d923f, start color #ed5fa6, plus, 6, end color #ed5fa6, n describes a different sequence.

In order to bring the formula 2, plus, 6, left parenthesis, n, minus, 1, right parenthesis to an equivalent formula of the form A, plus, B, n, we can expand the parentheses and simplify:
\begin{aligned}&\phantom{=}2+6(n-1)\\\\ &=2+6n-6\\\\ &=-4+6n\end{aligned}
Some people might prefer the formula minus, 4, plus, 6, n over the equivalent formula 2, plus, 6, left parenthesis, n, minus, 1, right parenthesis, because it’s shorter. The nice thing about the longer formula is that it gives us the first term.

Explicit Equation for Exponential Growth

Explicit Formula
Explicit Formula
A Population Grows According to An Exponential Growth Model. The Initial Populations P Sub 0 = 16 And The Growth Rate Is R = 0.35. Which Means The Growth Rate would Be 35%. So Every End Unit Of Time the Population Grows Or Increases By 35%.
But For The Recursive And explicit Formulas Given Here, We Do Need To Use The Decimal form Of The Growth Rate. So For The Recursive Formula We Would Have P Sub N= The Quantity 1 + R That Would Be 1 + 0.35 Sowed Have 1.35 X P Sub And – 1.
So Now When We Find P Sub 1,n = 1, So I’ll Substitute 1 Whereas Well As Here. So P Sub 1 Is = To 1.35x P Sub 1 – 1 That Would Be P Sub Which We Know Is Equal To 16. So 1.35 X 16 = 21.6.so P Sub 1 = 21.6. Now When We Find P Sub 2,n = 2. So We’ll Have 1.35 X P Sub 2 – that’s P Sub 1, Which We Just Found. That’s Equal To 21.6. And Now We’ll Go Back to The Calculator 1.35 X 21.6 = 29.16. Notice When Using the Recursive Formula To Find The Next Value Of P We Always Use The Previous value Of P.
Next We’re Asked To Find The explicit Formula For P Sub N, Which Is Given Hereford Exponential Growth. P Sub N = P Sub 0 X The Quantity + R Raise To The Power Of N. And We Already Have All the Information We Need. P Sub 0 = 16and R = 0.35.
So We Have P Sub N = P Sub which Is 16 X The Quantity 1 + R Which Is 1.35raise To The Power Of N. Now We Can Use Our Explicit formula To Find P Sub 11 Without Knowing P Sub 10. P Sub 11 Is = To 16x 1.35 Raise To The Power Of 11. So 16 X 1.35raise To The Power Of 11. Our Directions Do Say Round Toat Least 1 Decimal Place. So If Round To The Tenths, Notice How We Have A 0in The Hundredths So This Would Be approximately 434.3. I Hope You Found This Helpful.