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3 years ago

Find the square. (4x – 6y3)2

Mathematics

1 answer:

Artyom0805 [142]3 years ago

6 0

Answer:

16x^2 -48xy^3 +36y^6

Step-by-step explanation:

Use FOIL or the distributive property or the form of the square of a binomial to expand the square.

(4x -6y^3)(4x -6y^3) = 4x(4x -6y^3) -6y^3(4x -6y^3)

= (4x)^2 -(4x)(6y^3) -(6y^3)(4x) +(6y^3)^2

= 16x^2 -48xy^3 +36y^6

_____

The form referred to above is ...

(a +b)^2 = a^2 +2ab +b^2

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3 years ago

Read 2 more answers

a) What is an alternating series? An alternating series is a whose terms are__________ . (b) Under what conditions does an alter

andriy [413]

Answer:

a) An alternating series is a whose terms are alternately positive and negative

b) An alternating series $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n$ where bn = |an|, converges if $0< b_{n+1} \leq b_n$ for all n, and $\lim_{n \to \infty} b_n =$ 0

c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1

Step-by-step explanation:

Part a

An Alternating series is an infinite series given on these three possible general forms given by:

$\sum_{n=0}^{\infty} (-1)^{n} b_n$

$\sum_{n=0}^{\infty} (-1)^{n+1} b_n$

$\sum_{n=0}^{\infty} (-1)^{n-1} b_n$

For all $a_n >0, \forall n$

The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.

Part b

An alternating series $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n$ where bn = |an| converges if $0< b_{n+1} \leq b_n$ for all n and $\lim_{n \to \infty} b_n$ =0

Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:

Theorem (Alternating series test)

If a sequence of positive terms {bn} is monotonically decreasing and

$\lim_{n \to \infty} b_n = 0$ , then the alternating series $\sum (-1)^{n-1} b_n$ converges if:

i) $0 \leq b_{n+1} \leq b_n \forall n$

ii) $\lim_{n \to \infty} b_n = 0$

then $\sum_{n=1}^{\infty}(-1)^{n-1} b_n$ converges

Proof

For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then

$|s_n -s| < \epsilon$ for all n, and that implies that the series converges to a value, s.

And this complete the proof.

Part c

An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series

By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:

Rn = s - sn

Where $s_n$ represent the partial sum for the series and s the total for the sum.

Is important to notice that the size of the error is at most $b_{n+1}$ by the following theorem:

Theorem (Alternating series sum estimation)

If $\sum (-1)^{n-1} b_n$ is the sum of an alternating series that satisfies

i) $0 \leq b_{n+1} \leq b_n \forall n$

ii) $\lim_{n \to \infty} b_n = 0$

Then then $\mid s - s_n \mid \leq b_{n+1}$

Proof

In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any two consecutive partial sums sn and sn+1.

$\mid{s -s_n} \mid \leq \mid{s_{n+1} -s_n}\mid = b_{n+1}$