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i)expand (2+3x)^6 in arcending order power of x up to and including the term in x^2 simplifying the coefficient.

ii)given that the coefficient of x^2 in the expansion of (1+ax)(2+3x)^4  is 2304 fing the value of the constant a.

 Apr 14, 2022
 #1
avatar+64 
-2

yea i cant do this one 

Srry

-Kash😎😎😎

 Apr 14, 2022
 #2
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+1

why bother posting then

Guest Apr 14, 2022
 #3
avatar+580 
+3

i) Applybinomialtheorem:(a+b)n=βˆ‘ni=0(ni)a(nβˆ’i)bi

 

βˆ‘6i=0(6i)β‹…2(6βˆ’i)(3x)i

 

*After Plugging in Calc*

 

6!0!(6βˆ’0)!β‹…26(3x)0+6!1!(6βˆ’1)!β‹…25(3x)1+6!2!(6βˆ’2)!β‹…24(3x)2+6!3!(6βˆ’3)!β‹…23(3x)3+6!4!(6βˆ’4)!β‹…22(3x)4+6!5!(6βˆ’5)!β‹…21(3x)5+6!6!(6βˆ’6)!β‹…20(3x)6

64+576x+2160x2+4320x3+4860x4+2916x5+729x6

 Apr 14, 2022
 #4
avatar+580 
+1

I really don't know how to do ii)

Vinculum  Apr 14, 2022
edited by Vinculum  Apr 14, 2022

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