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12 Dec 2016, 06:51
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D
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Given that when expanding \(\big (x-\frac{1}{3} \big )^n\), the coefficient of 3rd term of the sum is 5. What is the mid term of the sum?
A. \(\frac{70}{240}x^4\)
B. \(\frac{28}{27}x^5\)
C. \(\frac{70}{27}x^6\)
D. \(\frac{-28}{27}x^5\)
E. \(\frac{-70}{27}x^6\)
A. \(\frac{70}{240}x^4\)
B. \(\frac{28}{27}x^5\)
C. \(\frac{70}{27}x^6\)
D. \(\frac{-28}{27}x^5\)
E. \(\frac{-70}{27}x^6\)
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12 Dec 2016, 07:21
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answer D give me a jiff to type all that
12 Dec 2016, 07:39
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nguyendinhtuong
Given that when expanding \(\big (x-\frac{1}{3} \big )^n\), the coefficient of 3rd term of the sum is 5. What is the mid term of the sum?
A. \(\frac{70}{240}x^4\)
B. \(\frac{28}{27}x^5\)
C. \(\frac{70}{27}x^6\)
D. \(\frac{-28}{27}x^5\)
E. \(\frac{-70}{27}x^6\)
A. \(\frac{70}{240}x^4\)
B. \(\frac{28}{27}x^5\)
C. \(\frac{70}{27}x^6\)
D. \(\frac{-28}{27}x^5\)
E. \(\frac{-70}{27}x^6\)
I’ll start from binomial theorem:
\((a + b)^n = a^n + nC1a^{n-1}*b + nC2a^{n-2}*b^2 …\)
Now we have: “the coefficient of 3rd term of the sum is 5”. We must keep in mind that our b actually divides coefficient and in order to find n we need to multiply by cancelled factor \((\frac{1}{b})^2\).
\(nC2 = 5*9\)
\(\frac{n!}{2*(n-2)!} = 45\)
Simplifying this:
\(n(n-1) = 90 = 10*9\)
Our \(n=10\)
Now back to binomial theorem.
If our power is \(10\) than we have:
\(x^{10} +10* x^9*(\frac{-1}{3}) + 45*x^8*\frac{(-1}{3})^2 +120* x^7*(\frac{-1}{3})^3 +210* x^6*(\frac{-1}{3})^4 + 252* x^5*(\frac{-1}{3})^5\) … I’ll stop here because 5th power is our middle term.
From that:
\(\frac{252*x^5}{(-243)} = \frac{- 28}{27} *x^5\)
Answer D.
