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Question: Sum of odd terms is *A* and sum of even terms is *B* in the expansion of \((x + a)^{n}\), then...

Sum of odd terms is A and sum of even terms is B in the expansion of (x+a)n(x + a)^{n}, then

A

AB=14(x−a)2n−(x+a)2nAB = \frac{1}{4}(x - a)^{2n} - (x + a)^{2n}

B

2AB=(x+a)2n−(x−a)2n2AB = (x + a)^{2n} - (x - a)^{2n}

C

4AB=(x+a)2n−(x−a)2n4AB = (x + a)^{2n} - (x - a)^{2n}

D

None of these

Answer

4AB=(x+a)2n−(x−a)2n4AB = (x + a)^{2n} - (x - a)^{2n}

Explanation

Solution

(x+a)n=n⥂C0xn+n⥂C1xn−1a1+n⥂C2xn−2a2+...+n⥂Cnxn−n.an=(xn+n⥂C2xn−2a2+..)+(n⥂C1xn−1a1+n⥂C3xn−3a3+....)=A+B(x + a)^{n} =^{n} ⥂ C_{0}x^{n} +^{n} ⥂ C_{1}x^{n - 1}a^{1} +^{n} ⥂ C_{2}x^{n - 2}a^{2} + ... +^{n} ⥂ C_{n}x^{n - n}.a^{n} = (x^{n} +^{n} ⥂ C_{2}x^{n - 2}a^{2} + ..) + (^{n} ⥂ C_{1}x^{n - 1}a^{1} +^{n} ⥂ C_{3}x^{n - 3}a^{3} + ....) = A + B.....(i)

Similarly, (x−a)n=A−B(x - a)^{n} = A - B .....(ii)

From (i) and (ii), we get 4AB=(x+a)2n−(x−a)2n4AB = (x + a)^{2n} - (x - a)^{2n}

Trick: Put n=1n = 1 in (x+a)n(x + a)^{n}. Then, x+a=A+Bx + a = A + B.

Comparing both sides A=x,B=aA = x,B = a.

Option (3) L.H.S. 4AB=4xa4AB = 4xa, R.H.S. (x+a)2−(x−a)2=4ax(x + a)^{2} - (x - a)^{2} = 4ax. i.e., L.H.S. = R.H.S