Write the first four terms in the binomial expansion of ( 1 + x/2)to the power of 9
I would use Pascal's triangle. Ignore the 1/2 in the second term, get the basic coefficents, then multiply each of the four terms with (1/2)^(n-1) for the first, second, third, and fourth term.
To find the first four terms in the binomial expansion of (1 + x/2)^9, you can use the binomial theorem. The binomial theorem states that for any two numbers a and b and any non-negative integer n, the expansion of (a + b)^n can be found using the formula:
(a + b)^n = nC0 * a^n * b^0 + nC1 * a^(n-1) * b^1 + nC2 * a^(n-2) * b^2 + ... + nCr * a^(n-r) * b^r + ... + nCn * a^0 * b^n
where nCr represents the combination of n items taken r at a time, and can be calculated as:
nCr = n! / (r! * (n-r)!)
Now let's apply this formula to find the first four terms in the expansion of (1 + x/2)^9:
n = 9, a = 1, b = x/2
Term 1 (nCr = 9C0):
nC0 * a^9 * b^0 = 1 * 1^9 * (x/2)^0 = 1
Term 2 (nCr = 9C1):
nC1 * a^8 * b^1 = 9 * 1^8 * (x/2)^1 = 9 * x/2 = (9/2) * x
Term 3 (nCr = 9C2):
nC2 * a^7 * b^2 = 9C2 * 1^7 * (x/2)^2 = 36 * x^2/4 = 9 * x^2/2 = (9/2) * x^2
Term 4 (nCr = 9C3):
nC3 * a^6 * b^3 = 9C3 * 1^6 * (x/2)^3 = 84 * x^3/8 = 21 * x^3/2
Therefore, the first four terms in the binomial expansion of (1 + x/2)^9 are:
1, (9/2) * x, (9/2) * x^2, (21/2) * x^3