Question

In triangle ABC, the value of $\sum_{r = 0}^{n}{nC_{r}}$a^r · b^n–r · cos

(rB – (n – r) A) is equal to –

• A. cⁿ
• B. Zero
• C. aⁿ
• D. bⁿ

Answer 1

Answer: (A)

cⁿ

Answer 2

$\sum_{r = 0}^{n}{nC_{r}}$a^r · b^n–r · cos (rB – (n – r) A)

= Re $\left( \sum_{r = 0}^{n}{nC_{r}a^{r} \cdot b^{n - r} \cdot e^{i(rB - (n - r)A)}} \right)$

= Re $\left( \sum_{r = 0}^{n}{nC_{r}(a \cdot e^{iB})^{r} \cdot (b \cdot e^{- iA})^{n - r}} \right)$

= Re (a e^iB + b . e^–iA)ⁿ

= Re (a cos B + ia sin B + b cos A – ib sin A)ⁿ

= (a cos B + b cos A)ⁿ = cⁿ ..