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Find the middle term in the expansion of `(3x-(x^3)/6)^9`

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To find the middle term in the expansion of \((3x - \frac{x^3}{6})^9\), we can follow these steps: ### Step 1: Identify the number of terms in the expansion The number of terms in the expansion of \((a + b)^n\) is given by \(n + 1\). Here, \(n = 9\), so the number of terms is: \[ n + 1 = 9 + 1 = 10 \] Since there are 10 terms, the middle terms will be the 5th and 6th terms. ...
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ALLEN-Solutions of Triangle & Binomial Theorem-Illustration
  1. Find (a) the coefficient of x^7 in the epansion of (ax^2+1/(bx))^11...

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  2. Find the number of rational terms in exponents of (9^(1//4)+8^(1//6))^...

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  3. Find the middle term in the expansion of (3x-(x^3)/6)^9

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  4. Find the term independent of x in the expansion of (sqrt(x/3)+((sqrt3)...

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  5. Find numerically greatest term is the expansion of (3-5x)^11 "when " x...

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  6. Find the term independent of x in the expansion off the following ex...

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  7. Prove that : ""^(25)C(10)+""^(24)C(10)+……..+""^(10)C(10)=""^(26)C(11)

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  8. A student is allowed to select at most n books from a collection of (...

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  9. Prove that (i) C(1)+2C(2)+3C(3)+……+nC(n)=n.2^(n-1) (ii) C(0)+(C(1)...

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  10. If (1+x)^n=underset(r=0)overset(n)C(r)x^r then prove that C(1)^2+2.C(2...

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  11. If (1 + x)^(n) = C(0) + C(1) x + C(2)x^(2) + C(3) x^(3)+ …+ C(n) x^(n...

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  12. Prove that (""^(2n)C(0))^2-(""^(2n)C(1))^2+(""^(2n)C(2))^2-.....+(-1)^...

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  13. Prove that : ""^(n)C(0).""^(2n)C(n)-""^(n)C(1).""^(2n-2)Cn(n)+""^(n)...

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  14. If (1+x)^n=C(0)C1c+C(2)x^2+…..+C(n)x^n then show that the sum of the p...

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  15. If (1+x)^n=C(0)+C(1)x+C(2)x^2+….+C(n)x^n then prove that (SigmaSigma)...

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  16. Find the coffiecient of x^2 y^3 z^4 w in the expansion of (x-y-z+w)^(...

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  17. Find the total number of terms in the expansion of 1(1+x+y)^(10) and c...

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  18. Find the coffiecient of x^5 in the expansion of (2-x+ 3x ^2)^6

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  19. If (1+x+x^2)^n = underset(2n)overset(r=0)Sigma a (r)x^r then prove th...

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  20. If f ( x ) = [ x ] , where [ ⋅ ] denotes greatest integral function...

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