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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Binomial Theorem

BITSAT 2014 · Mathematics · Binomial Theorem

If T_ 0 , ~T_ 1 , ~T_ 2 T_ n represent the terms in the expansion...

If T0, T1, T2……Tn\mathrm{T}_{0}, \mathrm{~T}_{1}, \mathrm{~T}_{2} \ldots \ldots \mathrm{T}_{\mathrm{n}} represent the terms in the expansion of (x+a)n(x+a)^{n}, then \left(\mathrm{T}_{0}-\mathrm{T}_{2}+\mathrm{T}_{4}-\ldots \ldots .\right)^{2}+$\left(\mathrm{T}_{1}-\mathrm{T}_{3}+\mathrm{T}_{5}-\ldots . .\right)^{2}=

Options

  1. A(x2+a2)\left(x^{2}+a^{2}\right)
  2. B(x2+a2)n\left(x^{2}+a^{2}\right)^{n}
  3. C(x2+a2)1/n\left(x^{2}+a^{2}\right)^{1 / n}
  4. D(x2+a2)−1/n\left(x^{2}+a^{2}\right)^{-1 / n}

Correct answer

B. (x2+a2)n\left(x^{2}+a^{2}\right)^{n}

Step-by-step solution

From the given condition, replacing aa by a and −ai-a i respectively, we get (x+ai)n=(T0−T2+T4−……)+i(T1−T3+T5−….)...(i)(x+a i)^{n}= \left(T_{0}-T_{2}+T_{4}-\ldots \ldots\right)+i\left( T_{1}-T_{3}+T_{5}-\ldots .\right) ...(i) and (x-a i)^{n}= \left( T_{0}-T_{2}+T_{4}-\ldots \ldots\right)−i(T1−T3+T5−…)...(ii)Multiplying(ii)-i\left( T_{1}-T_{3}+T_{5}-\ldots\right)...(ii) Multiplying (ii) and (i) we get required result i.e., (x2+a2)n\left(x^{2}+a^{2}\right)^{n}=\left( T_{0}-T_{2}+T_{4}-\ldots .\right)^{2}+\left(T_{1}-T_{3}+T_{5}-\ldots\right)^{2}
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