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3 years ago

Can anyone explain how to expand and simplify 3(2a+5) +5 (a-2)

2 answers:

german3 years ago

8 0

First of all you need to expand this expression. you do this by multiplying the number outside the brackets with each term inside the brackets. this is the first one ; 3 x 2a = 6a, 3 x 5 = 15. the extra + sign in the first brackets stays between the two terms, so the first set of brackets is now 6a + 15. now we have to look at the second ; 5 x a = 5a, 5 x 2 = 10, like the extra + sign, the extra - sign stays between the two terms as well, so the second bracket is now 5a-10. you put them together to get 6a + 15 + 5a - 10.

next you need to simplify the expression. first you need to group like terms, e.g. the numbers and the letters. 6a + 15 +5a -10. the 6a + 5a = 11a and the 15 - 10 = +5. you then put these two terms together to get your final answer of 

11a + 5

Step2247 [10]3 years ago

5 0

Take the first bracket and it's multiplication, 3(2a+5)
You expand my multiplying everything inside the bracket by the number outside.
3*2a = 6a
3*5 = 15
Then multiply out the second bracket, 5(a-2)
5*a = 5a
5*-2 = -10
So now you have 6a + 15 + 5a - 10
Simplify it by collecting like terms, and you have:
11a + 5
And that's the answer, hope I helped!

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If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

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$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

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Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

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