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nCr?

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How to use nCr to find the 4th term of (3c+4d)^7

Mar 4, 2019

#1
+109519
+2

How to use nCr to find the 4th term of (3c+4d)^7

(3c+4d)^7

I know this is common wording for a question like this but I never know what end of the expansion that they want you to start from

anyway...

From one end it is

\(^7C_3\; (3c)^3(4d)^{(7-3)}\\\)

I used 3 because the first term is the 0th term.    So r=3 is the 4th term

If you started at the other end it would be

\(^7C_3\; (4d)^3(3c)^{(7-3)}\\\)

I suppose there is some convention about what end you are supposed to start at but i don't know what it is.

Mar 4, 2019
#2
+4569
+4

Melody is right. This is called the binomial theorem, and it's pretty helpful to find coefficients and constants in these types of expressions.

Starting from the first term, we have \(\binom{7}{0}(3c)^7(4d)^0+\binom{7}{1}(3c)^6(4d)^1+\binom{7}{2}(3c)^5(4d)^2+...+\binom{7}{7}(3c)^0(4d)^7\)

Thus, the fourth term should be \(\binom{7}{3}(3c)^4(4d)^3\) .

Fixed.

Mar 4, 2019
edited by tertre  Mar 4, 2019
#3
+109519
+1

Thanks Tertre but you put the powers for the first brackets in the wrong place.

It is just a careless error but it could be confusing to the asker.

Melody  Mar 4, 2019
#4
+4569
+2

Fixed, thanks Melody.

tertre  Mar 5, 2019