# Thread: [SOLVED] Question regarging Binomial Theorem

1. ## [SOLVED] Question regarging Binomial Theorem

Find n so that the coefficients of the eleventh and thirteenth terms in the expansion of (1+x)^n are the same.

I need some help regarding this question, any help would be greatly appreciated.

2. Originally Posted by Solid8Snake
Find n so that the coefficients of the eleventh and thirteenth terms in the expansion of (1+x)^n are the same.

I need some help regarding this question, any help would be greatly appreciated.
the expansion will have 23 terms.

what value of n will yield 23 terms?

hint : look for a pattern in the symmetry and the number of coefficients in the rows of Pascal's triangle ...

3. Hello, Solid8Snake!

Another approach . . .

Find $\displaystyle n$ so that the coefficients of the $\displaystyle 11^{th}$ and $\displaystyle 13^{th}$ terms in the expansion of $\displaystyle (1+x)^n$ are equal.
The coefficients of the 11th and 13th terms are: .$\displaystyle _nC_{10}\,\text { and }\,_nC_{12}$

We have: .$\displaystyle _nC_{10} \:=\:_nC_{12} \quad\Rightarrow\quad \frac{n!}{10!(n-10)!} \;=\;\frac{n!}{12!(n-12)!} \quad\Rightarrow\quad 10!(n-10)! \;=\;12!(n-12)!$

Divide by $\displaystyle 10!\!:\quad (n-10)! \;=\;12\cdot11\cdot(n-12)!$

Divide by $\displaystyle (n-12)!\!:\quad (n-10)(n-11) \;=\;12\cdot11 \quad\Rightarrow\quad n^2 - 21n - 22 \;=\;0$

. . $\displaystyle (n+1)(n-22) \;=\;0 \quad\Rightarrow\quad n \;=\;-1,22$

Therefore: .$\displaystyle n \:=\:22$