Question

binomial theorem: determine third term in expansion of (x^2 + 2x + 1 ) ^6

Answer 1

Try factoring x^2 + 2x + 1

Answer 2

i.e. write it as \(((x+1)^2)^6\)

Answer 3

giving you
\[(x+1)^{12}\] and expand that one

Answer 4

what about a permtatio/ combinatio formula?

Answer 5

like nCr or somethig like that..

Answer 6

yeah, something like that

Answer 7

can someone answer the question using nCr which prevents you from having to do the long factoring?

Answer 8

first term is \(x^{12}\)
second term is \(12x^{11}\)

Answer 9

\[\left(\begin{matrix}13 \\ 2\end{matrix}\right)x^{11}1^2\]

Answer 10

third term is \(\binom{12}{2}x^{10}\)

Answer 11

oops, that's suppose to be 12

Answer 12

and \(\binom{12}{2}\) is \(\frac{12\times 11}{2}=6\times 11=66\)

Answer 13

you might know \(\binom{12}{2}\) as \(_{12}C_2\) same thing

Answer 14

\[ncr=\frac{ n \left( n-1 \right)\left( n-2 \right)...\left( n-\left( r-1 \right) \right) }{ r*\left( r-1 \right)\left( r-2 \right)....3*2*1 }\]