To find the common difference, subtract the 1st term from the 2nd term
1 8/9 - 2 =
17/9 - 2 =
17/9 - 18/9 =
-1/9 <==
<span>binomial </span>is an algebraic expression containing 2 terms. For example, (x + y) is a binomial.
We sometimes need to expand binomials as follows:
(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
<span>(a + b)4</span> <span>= a4 + 4a3b</span><span> + 6a2b2 + 4ab3 + b4</span>
<span>(a + b)5</span> <span>= a5 + 5a4b</span> <span>+ 10a3b2</span><span> + 10a2b3 + 5ab4 + b5</span>
Clearly, doing this by direct multiplication gets quite tedious and can be rather difficult for larger powers or more complicated expressions.
Pascal's Triangle
We note that the coefficients (the numbers in front of each term) follow a pattern. [This was noticed long before Pascal, by the Chinese.]
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
You can use this pattern to form the coefficients, rather than multiply everything out as we did above.
The Binomial Theorem
We use the binomial theorem to help us expand binomials to any given power without direct multiplication. As we have seen, multiplication can be time-consuming or even not possible in some cases.
<span>Properties of the Binomial Expansion <span>(a + b)n</span></span><span><span>There are <span>\displaystyle{n}+{1}<span>n+1</span></span> terms.</span><span>The first term is <span>an</span> and the final term is <span>bn</span>.</span></span><span>Progressing from the first term to the last, the exponent of a decreases by <span>\displaystyle{1}1</span> from term to term while the exponent of b increases by <span>\displaystyle{1}1</span>. In addition, the sum of the exponents of a and b in each term is n.</span><span>If the coefficient of each term is multiplied by the exponent of a in that term, and the product is divided by the number of that term, we obtain the coefficient of the next term.</span>
The correct answer is A square root of 41
Answer:
Step-by-step explanation:
Assuming there is a punitive removal of one point for an incorrect response.
Five undiscernable choices: 20% chance of guessing correctly -- Expectation: 0.20*(1) + 0.80*(-1) = -0.60
Four undiscernable choices: 25% chance of guessing correctly -- Expectation: 0.25*(1) + 0.75*(-1) = -0.50
I'll use 0.33 as an approzimation for 1/3
Three undiscernable choices: 33% chance of guessing correctly -- Expectation: 0.33*(1) + 0.67*(-1) = -0.33 <== The approximation is a little ugly.
Two undiscernable choices: 50% chance of guessing correctly -- Expectation: 0.50*(1) + 0.50*(-1) = 0.00
And thus we see that only if you can remove three is guessing neutral. There is no time when guessing is advantageous.
One Correct Answer: 100% chance of guessing correctly -- Expectation: 1.00*(1) + 0.00*(-1) = 1.00
