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MISCELLANEOUS EXERCISE 4 (II) · Q120

Q.Show that there is no constant term in the expansion of (2x−x24)9\left(2x-\dfrac{x^2}{4}\right)^9.

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tr+1=9Cr(2x)9−r(−x24)r=9Cr29−r(−14)rx9−r+2r=9Cr29−r(−14)rx9+rt_{r+1}={}^9C_r(2x)^{9-r}\left(-\dfrac{x^2}4\right)^r={}^9C_r2^{9-r}\left(-\dfrac14\right)^rx^{9-r+2r}={}^9C_r2^{9-r}\left(-\dfrac14\right)^rx^{9+r}. For a constant term we would need 9+r=09+r=0, i.e. r=−9r=-9, which is not a valid value of rr (must be 0≤r≤90\le r\le9). Since the exponent 9+r9+r is at least 99 for …

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