Consider the expansion of (x + 2)11.
(a) Write down the number of terms in this expansion.
(b) Find the term containing x2.
(a) Well, to determine the number of terms in the expansion, we can use the binomial theorem! According to the theorem, the number of terms in the expansion of (x + 2)11 is equal to the power of the binomial plus one. So, in this case, there will be 11 + 1 = 12 terms in the expansion.
(b) Now, to find the term containing x^2, we can rely on the pattern observed in the binomial coefficient. In the expansion, the term containing x^2 will have a binomial coefficient of C(11, 2).
But wait, let's not get too technical! Instead of going through all those calculations, let me put a clownish spin on it: Imagine x is a sneaky little squirrel and (x + 2) is a tree. Now, we want to find the term where the squirrel jumps onto a branch that's 2 levels above the ground. So, when the squirrel performs its gymnastics, it lands on the branch that corresponds to x^2.
So, in the expansion of (x + 2)11, the term containing x^2 is C(11, 2).
To expand (x + 2)^11, we can use the binomial theorem. The binomial theorem states that the expansion of (a + b)^n can be written as the sum of the terms: (nC0)a^n b^0 + (nC1)a^(n-1) b^1 + ... + (nCr)a^(n-r) b^r + ... + (nCn)a^0 b^n, where nCr represents the combination formula nCr = n! / (r!(n-r)!).
(a) The number of terms in the expansion of (x + 2)^11 is 11 + 1 = 12 because the powers of a and b range from n to 0, in this case, from 11 to 0.
(b) To find the term containing x^2, we need to determine the value of r in the expansion. Using the formula nCr, we know that r is the power of a or the exponent on x. In this case, r = 2. So, the term containing x^2 is given by: (nC2)a^(11-2) b^2 = (11C2)x^2 2^(11-2).
To calculate (11C2), we use the combination formula:
(11C2) = 11! / (2!(11-2)!) = 11! / (2!9!) = (11 × 10) / (2 × 1) = 55.
Therefore, the term containing x^2 is 55x^2 2^9 or 55x^2 512.
To answer this question, we need to understand the binomial expansion formula. The expansion of (x + 2)^n can be found using the binomial theorem.
The binomial theorem states that:
(x + y)^n = nC0 * x^n * y^0 + nC1 * x^(n-1) * y^1 + nC2 * x^(n-2) * y^2 + ... + nCr * x^(n-r) * y^r + ... + nCn * x^0 * y^n
In this case, we have (x + 2)^11. So, we can substitute x as x and y as 2 in the binomial expansion formula.
(a) The number of terms in the expansion can be found using the binomial coefficient or combination formula, nCr. In this case, n is 11, and we want to find the term containing x^2. Using the binomial coefficient formula:
nC(r-1) = nCr = n! / [(n-r)! * r!]
We want to find the term containing x^2, which means the exponent of x is 2. Let's substitute that value in the formula:
11C2 = 11! / [(11-2)! * 2!]
Simplifying this expression, we get:
11C2 = 11! / 9! * 2!
Since 9! * 2! can be further simplified, we have:
11C2 = 11 * 10 / 2
Therefore, the number of terms in the expansion is 11 * 10 / 2 = 55.
(b) To find the term containing x^2, we need to identify the corresponding term in the expansion. According to the binomial theorem, the rth term in the expansion is given by:
nCr * x^(n-r) * y^r
In this case, we have n = 11 and r is the term number we are interested in. Since we want to find the term containing x^2, r would be 2.
Plugging in the values, the term containing x^2 can be written as:
11C2 * x^(11-2) * (2)^2
Substituting the values:
11C2 * x^9 * 4
We have already found the value of 11C2 earlier, which is 55. So, the term containing x^2 would be:
55 * x^9 * 4
Simplifying this expression, we get:
220 * x^9
Therefore, the term containing x^2 in the expansion of (x + 2)^11 is 220 * x^9.
I am sure you meant: (x + 2)^11
there will be 12 terms in the expansion.
they are:
x^11 + V(11,1) (x^10)(2) + C(11,2) (x^9)(2^2) + ... + C(11,9) x^2 (2^9) + C(11,10) x^1 (2^10) + 2^11
which term contains x^2 ??
calculate it.