Question

Any complex number in the polar form can be expressed in Euler's form as $\cos \theta +i\sin \theta =e^{i\theta }$. This form of the complex number is useful in finding the sum of series $\sum _{r=0}^n{}^n{C}_r(\cos \theta +i\sin \theta {)}^r$.
$\sum _{r=0}^n{}^n{C}_r(\cos r\theta +i\sin r\theta )=\sum _{r=0}^n{}^n{C}_r{e}^{ir\theta }=\sum _{r=0}^n{}^n{C}_r(e^{i\theta }{)}^r=(1+e^{i\theta }{)}^n$
Also, we know that the sum of binomial series does not change if $r$ is replaced by $n-r$. Using these facts, answer the following questions.

In triangle $ABC$, the value of $\sum _{r=0}^{50}{}^{50}{C}_r{a}^r{b}^{50-r}\cos (rB-(50-r\left)A\right)$ is equal to (where $a,b,c$ are sides of triangle opposite to angle $A,B,C$, respectively, and $s$ is semi-perimeter)

• A. $c^{49}$
• B. $(a+b{)}^{50}$
• C. $(2s-a-b{)}^{50}$
• D. None of these

Answer 1

The correct option is C $(2s-a-b{)}^{50}$

$\sum _{r=0}^{50}{}^{50}{C}_r{a}^r{b}^{50-r}\cos (rB-(50-r\left)A\right)$
$=\text{Re}\left(\sum _{r=0}^{50}{}^{50}{C}_r{a}^r{b}^{50-r}{e}^{i(rB-(50-r\left)A\right)}\right)$

$=\text{Re}(\sum _{r=0}^{50}{}^{50}{C}_r{(a\times e^{iB})}^r\times {(b\times e^{-iA})}^{50-r})$

$=\text{Re}{(a{e}^{iB}+b{e}^{-iA})}^{50}$
$=\text{Re}{(a\cos B+ia\sin B+b\cos A-ib\sin A)}^{50}$
$=\text{Re}{(a\cos B+b\cos A)}^{50}[\because a\sin B=b\sin A]$
$=c^{50}=(2s-a-b{)}^{50}$