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if the constant term of the binomial expansion is (2x -1/x)^n is -160 then what is the value of n?

Profile image of vedant  wasnik
13 Years agoGrade
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1 Answer

Profile image of Sher Mohammad
12 Years ago
kth term of the expansion is
nCk 2k*xk*(-1)n-k*xk-nconstant term so k=n-k =>k=n/2

now nCn/2 2n/2=160 and n/2 must be odd
for n=6 the equation holds.

Sher Mohammad
faculty askiitians