How do you find the third term of #(x+3)^12#?

2 Answers
Jul 31, 2016

Answer:

#3^(rd)# term of #(x+3)^12# is #594x^10#

Explanation:

Any binomial expansion of the type #(x+a)^n# has #n+1# terms starting from #0^(th)# term, and the #n^(th)# term is given by #""^nC_rx^(n-r)a^r#, where #""^nC_r=(n!)/((n-r)!r!)#.

Hence #3^(rd)# term of #(x+3)^12# will be with #n=2#

#""^12C_2x^(12-2)a^2#

= #(12!)/(10!2!)x^10a^2#

= #(12xx11xx10xx9xx8xx7xx6xx5xx4xx3xx2xx1)/((10xx9xx8xx7xx6xx5xx4xx3xx2xx1)(2xx1))x^10a^2#

= #(12xx11xxcancel(10xx9xx8xx7xx6xx5xx4xx3xx2xx1))/(cancel(10xx9xx8xx7xx6xx5xx4xx3xx2xx1)(2xx1))x^10a^2#

= #(12xx11)/(2xx1)x^10a^2#

= #6xx11xx x^10a^2#

= #66x^10a^2#

Hence #3^(rd)# term of #(x+3)^12# will be #66xxx^10xx3^2=594x^10#

Jul 31, 2016

Answer:

#594x^10#

Explanation:

By the binomial theorem

#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k#

where #((n),(k)) = (n!)/((n-k)!k!)#

In our example #a=x#, #b=3# and #n=12#

The first, second and third terms of #(x+3)^12# are:

#((12),(0))x^12 = x^12#

#((12),(1))3x^11 = 12/1*3x^11 = 36x^11#

#((12),(2))3^2x^10 = (12*11)/(2*1)*9x^10 = 66*9x^10 = 594x^10#

If you were looking for more or later terms in the expansion, then it might be easier to pick them out from the appropriate row of Pascal's triangle. Some people call the first row of Pascal's triangle the "#0#th", which would make this the "#12#th" row. Others (like myself) call it the #13#th row. Let's just call it the row that starts #1, 12#...

enter image source here

The terms on this row of Pascal's triangle are:

#((12),(0))#, #((12),(1))#, #((12),(2))#, ... #((12),(12))#.

So for example, the middle term of #(x+3)^12# is #924*3^6x^6 = 673596x^6#