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$(3x-1)^4=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$

Value required to be found :- $a_4+3a_3+9a_2+27a_1+81a_0$

I can find the value of $a_4+a_3+a_2+a_1+a_0$.Then I don't know how to continue.Please help.

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  • $\begingroup$ Can't you just expand it? $\endgroup$ – principal-ideal-domain Aug 20 '15 at 9:21
  • $\begingroup$ @principal-ideal-domain-Method looks dull...:-) $\endgroup$ – tatan Aug 20 '15 at 9:22
  • $\begingroup$ Sure, but I guess you would be faster by doing so than typing the question here. $\endgroup$ – principal-ideal-domain Aug 20 '15 at 9:24
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HINT: try putting $x=\frac 13$

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  • $\begingroup$ Nice @davidquinn. Your answers are always very intuitive and simple. $\endgroup$ – Shailesh Aug 20 '15 at 10:53
  • $\begingroup$ @Shailesh thanks very much! The length of my answers is usually in proportion to my skill and patience with Mathjax $\endgroup$ – David Quinn Aug 20 '15 at 11:53
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Hint: using Pascal's triangle/the binomial theorem you can easily expand $(3x-1)^4$ and determine $a_0,\dots,a_4$.

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