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JEE Main2012MathematicsBinomial TheoremActual

The number of terms in the expansion of (y^ 1 / 5 +x^ 1 / 10 )⁵⁵ , in which powers of x and y are free from radical signs are

Options

  1. Asix
  2. Btwelve
  3. Cseven
  4. Dfive

Correct answer

A. six

Step-by-step solution

Given expansion is (y^ 1 / 5 +x^ 1 / 10 )⁵⁵ The general term is T_ r+1 = ⁵⁵ C _r (y^ 1 / 5 )^ 55-r (x^ 1 10 )^r T_ r+1 would free from radical sign if powers of y and x are integers. i.e. 55-r 5 and r 10 are integer. r is multiple of 10 . Hence, r=0,10,20,30,40,50 It is an A.P. Thus, 50=0+(k-1) 1050=10 k-10 k=6 Thus, the six terms of the given expansion in which x and y are free from radical signs.

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