If T0,T1,⋯,Tn represent the terms in the expansion of (x+a)n, then the value of (T0−T2+T4−⋯)2+(T1−T3+T5−⋯)2 is
A
a2n
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B
(x+a)2n
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C
(x2+a2)n
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D
a2n
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Solution
The correct option is C(x2+a2)n We now that, (x+a)n=xn+C1xn−1a+C2xn−2a2+⋯=T0+T1+T2+T3+⋯=(T0+T2+T4+⋯)+(T1+T3+T5+⋯)
The 1st bracket involves even powers of a and 2nd bracket involves odd powers of a.
We know that, i2=−1,i4=1,i3=−i,i5=i
Therefore, replacing a by ai (x+ai)n=(T0−T2+T4−⋯)+i(T1−T3+T5−⋯)⋯(i)
Now, a→−ai (x−ai)n=(T0−T2+T4−⋯)−i(T1−T3+T5−⋯)⋯(ii) Multiplying (i) and (ii) (x2+a2)n=(T0−T2+T4−⋯)2+(T1−T3+T5−⋯)2