MATH PLS HELPP BRAINLIEST U KNO

Question 4

IRISSAK [1]

Answer:The height of the plant at the end of the end of the experiment. So it is the second one.

Step-by-step explanation:

7 0

2 years ago

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Sum to n terms of each of following series. (a) 1 - 7a + 13a ^ 2 - 19a ^ 3+...

julia-pushkina [17]

Notice that the difference in the absolute values of consecutive coefficients is constant:

|-7| - 1 = 6

13 - |-7| = 6

|-19| - 13 = 6

and so on. This means the coefficients in the given series

$\displaystyle \sum_{i=1}^\infty c_i a^{i-1} = \sum_{i=1}^\infty |c_i| (-a)^{i-1} = 1 - 7a + 13a^2 - 19a^3 + \cdots$

occur in arithmetic progression; in particular, we have first value $c_1 = 1$ and for $n>1$ , $|c_i|=|c_{i-1}|+6$ . Solving this recurrence, we end up with

$|c_i| = |c_1| + 6(i-1) \implies |c_i| = 6i - 5$

So, the sum to $n$ terms of this series is

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \underbrace{\sum_{i=1}^n i (-a)^{i-1}}_{S'} - 5 \underbrace{\sum_{i=1}^n (-a)^{i-1}}_S$

The second sum $S$ is a standard geometric series, which is easy to compute:

$S = 1 - a + a^2 - a^3 + \cdots + (-a)^{n-1}$

Multiply both sides by $-a$ :

$-aS = -a + a^2 - a^3 + a^4 - \cdots + (-a)^n$

Subtract this from $S$ to eliminate the intermediate terms to end up with

$S - (-aS) = 1 - (-a)^n \implies (1-(-a)) S = 1 - (-a)^n \implies S = \dfrac{1 - (-a)^n}{1 + a}$

The first sum $S'$ can be handled with simple algebraic manipulation.

$S' = \displaystyle \sum_{i=1}^n i (-a)^{i-1}$

$\displaystyle S' = \sum_{i=0}^{n-1} (i+1) (-a)^i$

$\displaystyle S' = \sum_{i=0}^{n-1} i (-a)^i + \sum_{i=0}^{n-1} (-a)^i$

$\displaystyle S' = \sum_{i=1}^{n-1} i (-a)^i + \sum_{i=1}^n (-a)^{i-1}$

$\displaystyle S' = \sum_{i=1}^n i (-a)^i - n (-a)^n + S$

$\displaystyle S' = -a \sum_{i=1}^n i (-a)^{i-1} - n (-a)^n + S$

$\displaystyle S' = -a S' - n (-a)^n + \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle (1 + a) S' = \dfrac{1 - (-a)^n - n (1 + a) (-a)^n}{1 + a}$

$\displaystyle S' = \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2}$

Putting everything together, we have

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 S' - 5 S$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2} - 5 \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} =\boxed{\dfrac{1 - 5a - (6n+1) (-a)^n + (6n-5) (-a)^{n+1}}{(1+a)^2}}$

8 0

2 years ago

Write three expressions that are equivalent to 70/-5

Darya [45]

Answer:

See below in bold.

Step-by-step explanation:

By division 70 / -5 = -14.

Multiplying 70/-5 by 2/2 = 140/-10.

A third would be 70 / ( 5 - 10).

6 0

3 years ago

A. 1 B. 0.89 C. -.89 D. -1

DaniilM [7]

The answer is b because it goes down .89 every time

5 0

3 years ago

Solve the system. 2x + y = −3 −2y = 6 + 4x Write each equation in slope-intercept form.

BlackZzzverrR [31]

2x + y = -3
y = -2x - 3

-2y = 6 + 4x
y = -2x - 3

when solved,this system of equations will have infinite solutions because u r dealing with the same line

6 0

3 years ago

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