In two or more complete sentences, analyze how to find the third term in the expansion of (2x + y)4.
The first term in the binomial is "x2", the second term in "3", and the power n is 6, so, counting from0 to 6, the Binomial Theorem gives me:(x2 + 3)6 = 6C0 (x2)6(3)0 + 6C1(x2)5(3)1 + 6C2 (x2)4(3)2 + 6C3 (x2)3(3)3+ 6C4 (x2)2(3)4 + 6C5 (x2)1(3)5 + 6C6 (x2)0(3)6Then simplifying gives me(1)(x12)(1) + (6)(x10)(3) + (15)(x8)(9) + (20)(x6)(27)+ (15)(x4)(81) + (6)(x2)(243) + (1)(1)(729)= x12 + 18x10 + 135x8 + 540x6 + 1215x4 + 1458x2 + 729
Answer:
The third term in the expansion of the given expression us:
\(24x^2y^2\)
Step-by-step explanation:
We are given an expression as:
\((2x+y)^4\)
We know that by using the binomial theorem the expansion of the expression of the type:
\((ax+by)^n\)
is given by:
\((ax+by)^n=n_C_0 (ax)^0(by)^{n-0}+n_C_1 (ax)^1(by)^{n-1}+...........+n_C_n (ax)^n(by)^{n-n}\)
This means that there are n+1 terms in the expansion of the type: \((ax+by)^n\)
such that the rth term is:
\(n_C_{r-1}\times (ax)^{r-1}\times (by)^{n-(r-1)}\)
Here we have:
n=4,a=2 and b=1
Now, the third term in the expansion of the given expression is:
\(4_C_2(2x)^2(y)^{4-2}\)
\(=\dfrac{4!}{2!\times 2!}\times 4x^2\times y^2\\\\\\=24x^2y^2\)