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3 years ago

14

Simplifypls help me ASAP

1 answer:

Strike441 [17]3 years ago

3 0

$\dfrac{\left(\dfrac{4}{3}ab^3\right)\left(\dfrac{6}{2}a^4\right)}{a^5}=\dfrac{\left(\dfrac{\not4^2}{\not3_1}\cdot\dfrac{\not6^2}{\not2_1}\right)(aa^4)(b^3)}{a^5}=\dfrac{4a^5b^3}{a^5}=\boxed{4b^3}$

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The recursive function $f(0) = 1$ , $f(n) = f(n-1) + 2n$ represents the nth term of a sequence. Determine the explicit function

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Step-by-step explanation:

Given

$f(0) = 1$

$f(n) = f(n-1) + 2n$

Required

Write an explicit formula

Let n = 1

$f(1) = f(1-1) + 2*1$

$f(1) = f(0) + 2$

$f(1) = 1 + 2 = 3$

Let n = 2

$f(2) = f(2-1) + 2*2$

$f(2) = f(1) + 4$

$f(2) = 3 + 4 = 7$

Let n =3

$f(3)=f(3-1) + 2 * 3$

$f(3)=f(2) + 6$

$f(3)=7+ 6 = 13$

Let n = 4

$f(4) = f(4 - 1) + 2 * 4$

$f(4) = f(3) + 8$

$f(4) = 13+ 8 = 21$

So, we have:

$f(1) = 3= 1 + 2 * 1=1+0*1*2*1$

$f(2) = 7= 3 + 2 * 2=1+1*2+2*2$

$f(3) = 13= 7 + 2 * 3= 1+2*3+2*3$

$f(4) = 21 = 13 + 2 * 4= 1+3*4 +2 *4$

Following the above pattern:

$f(n) = 1 + (n - 1) * n + 2 * n$

$f(n) = 1 + n(n - 1) + 2n$

Open bracket

$f(n) = 1 + n^2 - n + 2n$

$f(n) = 1 + n^2 + n$

$f(n) = n^2 + n+1$

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