MCA entrance question

Question

In triangle $ABC$, the value of $\ \displaystyle \sum_{r=0}^n\ ^nC_ra^rb^{n-r}\cos(rB-(n-r)A)$ is equal to

(a) $c^n$

(b) $b^n$

(c) $a^n$

(d) $0$

I have no idea how to start doing it , if there are any properties please mention it I am preparing for MCA.

Answer 1

Consider n=1,

The equation reduces to:

bCos (-A) + aCosB

Considering the triangle to be acute, cos(-A) = cosA

The equation becomes,

bCosA + aCosB

We know that in any triangle,

aCosB+ bCosA = c

Substitute n=1, and we get answer A as c^n

Answer 2

The given expression equals $$\sum_{r=0}^{n}\binom{n}{r}a^rb^{n-r}(\frac{e^{i \theta} + e^{-i \theta}}{2})$$ where $\theta = rB - (n-r)A$.

$$\sum_{r=0}^{n}\binom{n}{r}(ae^{iB})^r \times (be^{-iA})^{n-r}$$ $$=(ae^{iB} + be^{-iA})^n=((a \cos B +b \cos A)+(a \sin B - b \sin A))^n$$ $$=(c + 0)^n$$

since $a \cos B + \cos C = c$ and $\frac{a}{\sin A} = \frac{b}{\sin B}$

Similarly $$\sum_{r=0}^{n}\binom{n}{r}(ae^{-iB})^r \times (be^{iA})^{n-r}=c^n$$

And hence the answer is $c^n$.