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\sf \: If (3x-1) ^(7) = a _(7) x ^(7) + a _(6) x ^(6) + a _(5) x ^(5)+.... \\ \sf + a₁x + a _(0), then \: \: a_0 + a₁ + a₂ + a_3 + a_4 + a_5 + a_6 + a7 =

a)0
b)1
c)128
d)64

Please solve this for me. Need step-by-step explanation. Spam free answer required. Thank you in anticipation.



2 Answers

2 votes

Answer:

hope it helps have a great day

Answer is option C

\sf \: If (3x-1) ^(7) = a _(7) x ^(7) + a _(6) x ^(6) + a _(5) x ^(5)+.... \\ \sf-example-1
User Anjie
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4.3k points
4 votes
Answer is c) 128

I added the step by step process in the pic attached, I don't know how to explain it very well but you can use binomial theorem for (3x-1)^7

Then you can add up all the coefficients of the calculated values as the question mentions afterwards a^7 + a^6 +..... which means it only wants you to add up the coefficients of the consecutive x terms, otherwise it would say a^7x^7 + a^6x^6 +..... and so on

Hope this helps :)
\sf \: If (3x-1) ^(7) = a _(7) x ^(7) + a _(6) x ^(6) + a _(5) x ^(5)+.... \\ \sf-example-1
User Jensen
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3.9k points