[SOLVED] Question regarging Binomial Theorem

**Solid8Snake** · September 3rd 2009, 02:55 PM

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.
Thx in advance!

**skeeter** · September 3rd 2009, 03:03 PM

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.
Thx in advance!

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 ...

**Soroban** · September 3rd 2009, 05:46 PM

Hello, Solid8Snake!

Another approach . . .

Find $n$ so that the coefficients of the $11^{th}$ and $13^{th}$ terms in the expansion of $(1+x)^n$ are equal.

The coefficients of the 11th and 13th terms are: . $_nC_{10}\,\text { and }\,_nC_{12}$

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

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

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

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

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

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