Find the 12th term in the expansion (5x+2)^14?

**rowdy3** · November 11th 2010, 06:33 PM

Find the 12th term in the expansion (5x+2)^14
I did (14 ncr 11) (5x)^3 (2)^11
364 (125x^3) (2048)
93,184,000x^3. Is that right?

**pickslides** · November 11th 2010, 06:37 PM

Originally Posted by rowdy3

I did (14 ncr 11) (5x)^3 (2)^11

Looks fine to me.

**tonio** · November 11th 2010, 07:42 PM

Originally Posted by rowdy3

Find the 12th term in the expansion (5x+2)^14
I did (14 ncr 11) (5x)^3 (2)^11
364 (125x^3) (2048)
93,184,000x^3. Is that right?

This can be tricky: you can both write $(5x+2)^{14}=\displaystyle{\sum\limits^{14}_{k=0}\b inom{14}{k}(5x)^{14-k}2^k=\sum\limits^{14}_{k=0}\binom{14}{k}(5x)^{k}2 ^{14-k}}$

In the first case you obtain $\displaystyle{\binom{14}{11}(5x)^32^{11}}$ , as you did, whereas in the second case you

get $\displaystyle{\binom{14}{11}(5x)^{11}2^3}$ , so what's the correct one?

In my opinion, they both are, and this perhaps is the reason why the usually ask what the

coefficient of some power of x is, not what is this or that term.

Tonio

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