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In the expansion of (2-x+3x^2)^6 the coefficient of x^5 is

MYQUESTIONS , 10 Years ago
Grade 11
anser 1 Answers
Arun Kumar

Last Activity: 10 Years ago

Hello Student,

The general term in expansion is
\\ {6! 2^r(-x)^s(3x^2)^t \over r!s!t!} \\={6! 2^r(-1)^s3^tx^{s+2t} \over r!s!t!} \\r+s+t=6 \\$for $x^5 \\s+2t=5 \\=>s=5-2t$ and $r=1+t \\=>t=0=>r=1,s=5 \\=>t=1=>r=2,s=3 \\=>t=2=>r=3,s=1 \\$three terms$ \\ {6! 2^1(-1)^53^tx^{5} \over 1!5!0!}+{6! 2^2(-1)^33^1x^{3+2} \over 2!3!1!}+{6! 2^3(-1)^13^2x^{1+2*2} \over 3!1!2!} \\=>-12-720-4320=-5052
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
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