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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+114 
-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+578 
+3

i) \(\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i\)

 

\(\sum _{i=0}^6\binom{6}{i}\cdot \:2^{\left(6-i\right)}\left(3x\right)^i\)

 

*After Plugging in Calc*

 

\(\frac{6!}{0!\left(6-0\right)!}\cdot \:2^6\left(3x\right)^0+\frac{6!}{1!\left(6-1\right)!}\cdot \:2^5\left(3x\right)^1+\frac{6!}{2!\left(6-2\right)!}\cdot \:2^4\left(3x\right)^2+\frac{6!}{3!\left(6-3\right)!}\cdot \:2^3\left(3x\right)^3+\frac{6!}{4!\left(6-4\right)!}\cdot \:2^2\left(3x\right)^4+\frac{6!}{5!\left(6-5\right)!}\cdot \:2^1\left(3x\right)^5+\frac{6!}{6!\left(6-6\right)!}\cdot \:2^0\left(3x\right)^6\)

\(64+576x+2160x^2+4320x^3+4860x^4+2916x^5+729x^6\)

 Apr 14, 2022
 #4
avatar+578 
+1

I really don't know how to do ii)

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

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