questions on attachment!!!!

- Jul 26th 2006, 06:56 PMLane10 questions that I need HELP ON!!! Please i can't do them!!!
questions on attachment!!!!

- Jul 26th 2006, 08:23 PMThePerfectHackerQuote:

Use sigma notation to write the sum. -2.4-.4+1.6+3.6

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Find a formula for an for the arithmetic sequence. a1=7, a11=19

$\displaystyle a_{11}=a_1+10k$

Substitute know values,

$\displaystyle 19=7+10k$

Thus,

$\displaystyle 12=10k$

Thus,

$\displaystyle k=1.2$

"An arithemtica sequnce with initial term of 7 and constant diffrence of 1.2"

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Find a formula for an for the arithmetic sequence. a1=11, a11=15

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A rubber ball on a hard surface takes a sequence of vertical bounces. Each bounce is 1/6 as high as the proceeding one. If this ball is dropped from a height of 12 feet, find the total distance it has traveled when it hits the surface the fifth time.

$\displaystyle 12+12(1/6)+12(1/6)(1/6)+12(1/6)(1/6)(1/6)+...$

Thus,

$\displaystyle 12+12(1/6)+12(1/6)^2+...+12(1/6)^4$---> Because you only got 5 bounces (do not think the exponent is 4 :eek: a common mistake it should be 4)

Thus, factor

$\displaystyle 12(1+(1/6)+...+(1/6)^4)$

Use sum for geometric series,

$\displaystyle 12\cdot \left( \frac{1-(1/6)^6}{1-1/6} \right)$

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6. Write the first five terms of the sequence.

a0=-2

a1=4

an=-2a n -1-4an-2

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7. Find the coefficient of b6a2 in the expansion of (b+2a)8

$\displaystyle {8 \choose 3}=56$

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8. Use the binomial theorem to expand and simplify the expression.

(s-u)5

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9. A college has seven instructors qualified to teach a special computer lab course which requires two instructors to be present. How many different pairs of teachers could teach the class.

$\displaystyle {7 \choose 2}=21$

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10. Eight cards are drawn, without replacement, from a standard deck of 52 cards. How many sets of eight cards are possible?

$\displaystyle {52 \choose 8}=752538150$

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11.

- Jul 27th 2006, 05:48 AMSoroban
Hello, Lane!

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Use sigma notation to write the sum: $\displaystyle -2.4 - 0.4+1.6+3.6$

Here are two possible sigma-representations.

It starts with $\displaystyle -2.4$ ... and 2 is added three times: .$\displaystyle \sum^3_{n=0}(-2.4 + 2n)$

It starts with $\displaystyle -4.4$ and 2 is added four times: .$\displaystyle \sum^4_{n=1}(-4.4 + 2n)$

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2. Find a formula for $\displaystyle a_n$ for the arithemtic sequence: $\displaystyle a_1 = 7,\;a_{11}=19$

Recall the formula for the $\displaystyle n^{th}$term: .$\displaystyle a_n\:=\:a_1 + d(n-1)$

. . where $\displaystyle a_1$ is the first term and $\displaystyle d$ is the common difference.

We are given $\displaystyle a_1=7$, so the formula becomes: .$\displaystyle a_n\:=\:7 + d(n-1)$

We are given $\displaystyle a_{11}=19$ ... the $\displaystyle 11^{th}$ term is $\displaystyle 19.$

So we have: .$\displaystyle 19\:=\:7 + d(11-1)\quad\Rightarrow\quad d = 1.2$

Therefore: .$\displaystyle a_n\:=\:7 + 1.2(n-1)$

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5. A rubber ball on a hard surface takes a sequence of vertical bounces.

Each bounce is 1/6 as high as the preceeding one.

If the ball is dropped from a height of 12 feet,

find the total distance it has traveled when it hits the surface for the fifth time.

**Warning**: "The Bouncing Ball" is a classic*trick question.*

I'll baby-step through the explanation . . .

The ball falls $\displaystyle 12$ feet. .(Contact #1)

It bounces up .$\displaystyle \frac{1}{6}(12)$ feet . . . and falls .$\displaystyle \frac{1}{6}(12)$ feet. .(Contact #2)

It bounces up $\displaystyle \left(\frac{1}{6}\right)^2(12)$ feet . . . and falls $\displaystyle \left(\frac{1}{6}\right)^2(12)$ feet. .(Contact #3)

. . You get the idea . . .

By the 5th contact, the total distance is:

. . $\displaystyle D \:=\:12 + 2\cdot12\left(\frac{1}{6}\right) + 2\cdot12\left(\frac{1}{6}\right)^2 +$$\displaystyle 2\cdot12\left(\frac{1}{6}\right)^3 + 2\cdot12\left(\frac{1}{6}\right)^4$

. . $\displaystyle D\:=\:12 + 4\bigg[1 + \frac{1}{6} + \left(\frac{1}{6}\right)^2 +$$\displaystyle \left(\frac{1}{6}\right)^3\bigg]$ $\displaystyle \:=\:12 + 4\left(\frac{259}{216}\right)\:=\:\boxed{\frac{907 }{54}\text{ feet}}$

- Jul 27th 2006, 06:50 AMSoroban
Hello again, Lane!

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7) Find the coefficient of $\displaystyle b^6a^2$ in the expansion of $\displaystyle (b+2a)^8$

The expansion begins: .$\displaystyle \binom{8}{0}b^8 + \binom{8}{1}b^7(2a) + \underbrace{\binom{8}{2}b^6(2a)^2} + \hdots$

The third term is: .$\displaystyle \frac{8!}{2!\,6!}\,b^6(4a^2) \;=\;\boxed{112}\,b^6a^2$

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8. Use the binomial theorem to expand and simplify: $\displaystyle (s - u)^5$

If you don't know the Binomial Theorem, we can't be much help.

$\displaystyle (s - u)^5\;=\;\binom{5}{0}s^5 + \binom{5}{1}s^4(-u) + \binom{5}{2}s^3(-u)^2 + \binom{5}{3}s^2(-u)^3$$\displaystyle + \binom{5}{4}s(-u)^4 + \binom{5}{5}(-u)^5$

. . I expanded it . . .can simplify it.*you*