InterviewSolution
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InterviewSolution
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Find the coffiecient of `x^5` in the expansion of `(2-x+ 3x ^2)^6` |
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Answer» The general term in the expansion of `(2-x+3x^2)^6=(6!)/(r !s! t!) 2 ^r(-x)^3(3x^2)^t` where r+s+ t =6 `(6!)/(r! s!t!)2 r xx (-1)xx(3)^t xxx ^(s+2t)` for the cofficient of `x^5` we must have s+2t =5 . `therefore S= 5-2 t and r=1 +t ` where ` 0 le r, s, t le 6 ` Now t=0 `rarr` r= 1 , s= 5 t=1 `rArr ` , s=3 , t = 2 `rArr` r= 3 , s=1 Thus there are three terms containig `x^5` and coefficient of `x^5` `=(6 !)/(1 ! 5 ! 0 ! )xx2^1xx(-1)^5xx3^0+(6 !)/(2! 3 !1 ! )xx2^2xx(-3)^1+(6!)/(3 ! 1 !2! ) xx2^3 xx (-1)^1 xx 3^2` ltbrge-12- 720 - 4320 = - 5052 |
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