Find the coefficient of the fourth term of (x+2)^5 •32 •48 •64 •80 find the coefficient of the third term of (3x-1)^5 •248 •270 •360 •505 find the coefficient of the third term of (a+5b^2)^4 •150 •500 •625 •1500

The coefficients of the binomial expansion , where n is the row number, is given in the Pascal's triangle shown below.

First, to find the coefficient of the fourth term of (x+2)^5 we look at row 5, term 4. The coefficient there is 10.

But, we must also remember that the term 2 also is taken to a certain power here. Mainly , for each term, the power of 2 is as follows:

2^0,  2^1,  2^2,  2^3=8.

So, in total we have: 10*8=80.

Second, to find  the coefficient of the third term of (3x-1)^5 we again go to the row 5, this time term 3 and we have 10 there. Now we must check how each of (3x) and 1 expand, now being careful about the sign as well.

we have:
                     (3x)^5 (1)         -(3x)^4 (1)         (3x)^3(1)=27x^3.

Thus, the coefficient of the third term is 27*10=270.

 
Third, we want to find the coefficient of the third term of (a+5b^2)^4. We look at row 4, term 3. There we have 6.

The terms a and 5b^2 are as follows:

             a^4 (5b^2)^0         a^3 (5b^2)^1       a^2 (5b^2)^2=25a^2b^4

Thus, the coefficient is 25*6=150.

80; 270; 150

Jim has run 5/6 of a mile so far. i hope that answers your question! good luck : )