First slide

Binomial theorem for positive integral Index

Question

Suppose $(x + a)^{n} = T_{0} + T_{1} + T_{2} + \dots + T_{n}$ Then ${(T_{0} - T_{2} + T_{4} - \dots)}^{2} + {(T_{1} - T_{3} + T_{5} - \dots)}^{2}$ is equal to

Moderate

A

${(x^{2} + a^{2})}^{n}$
B

${(x^{2} - a^{2})}^{n}$
C

$(x + a)^{2 n}$
D

$(x - a)^{2 n}$

Solution

Not that $T_{r} =^{n} C_{r} x^{n - r} a^{r} .$
$\begin{array}{l} (x + a i)^{n} =^{n} C_{0} x^{n} +^{n} C_{r} x^{n - 1} (i a) +^{n} C_{2} x^{n - 2} (i a)^{2} +^{n} C_{3} x^{n - 3} (i a)^{3} + \dots +^{n} C_{n} (i a)^{n} \\ = (T_{0} - T_{2} + T_{4} - \dots) + i ({\bar{T}}_{1} - {\bar{T}}_{3} + {\bar{T}}_{5} - \dots) \\ \begin{array}{l} \Rightarrow {(x^{2} + a^{2})}^{n} = {1 (x + a i)^{n}|}^{2} \\ = {(T_{0} - T_{2} + T_{4} - \dots)}^{2} + {(T_{1} - T_{3} + T_{5} - \dots)}^{2} \end{array} \end{array}$

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