Find the term independent of x in the expansion of (1+x+2x3)(3x22−13x)9
A
1754
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B
1752
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C
1954
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D
1952
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Solution
The correct option is A1754 For independent term in expansion (1+x+2x3)(3x22−13x)9 it can be written as (3x22−13x)9(1)+x(3x22−13x)9(2)+2x3(3x22−13x)9(3) Now solving each for independent term of x (r+1)th term for (3x22−13x)9 Tr+1=9Cr(32x2)9−r(−13x)r=9Cr(32)9−r(−13)rx18−3r Now for (1) to be independent fom x 18−3r=0⇒r=6 So 9Cr(32)9−r(−13)r=718 For (2) to be independent from x 18−3r=−1⇒r=193 Not possible For (3) to be independent from x 18−3r=−3⇒r=7 So 9Cr(32)9−r(−13)r=−127 Hence the required term will be 718−2⋅127=1754