The term independent of x in the expansion of
\((1−x^2+3x^3)(\frac{5}{2}x^3−\frac{1}{5x^2})11,x≠0 \)
is:
The term independent of x in the expansion of
\((1−x^2+3x^3)(\frac{5}{2}x^3−\frac{1}{5x^2})11,x≠0 \)
is:
\(\frac{7}{40}\)
\(\frac{33}{200}\)
\(\frac{39}{200}\)
\(\frac{11}{50}\)
The Correct Option is B
Solution and Explanation
The correct answer is (B) : \(\frac{33}{200}\)
\((1−x^2+3x^3)(\frac{5}{2}x^3−\frac{1}{5x^2})^{11},x≠0\)
General term of \((\frac{5}{2}x^3−\frac{1}{5x^2})^{11}\) is
\(T_{r+1}=^{11}C_r(\frac{5}{2}x^3)^{11−r}(\frac{−1}{5x^2})^r\)
\(=^{11}C_r(\frac{5}{2})^{11−r}(\frac{−1}{5})^r x^{33−5r}\)
So, term independent from x in given expression
\(=^{−11}C_7(\frac{5}{2})^4(\frac{−1}{5})^7=\frac{11×10×9×8}{24}×\frac{1}{16×125 }\)
\(=\frac{33}{200}\)
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Concepts Used:
Binomial Expansion Formula
The binomial expansion formula involves binomial coefficients which are of the form
(n/k)(or) nCk and it is calculated using the formula, nCk =n! / [(n - k)! k!]. The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.
This binomial expansion formula gives the expansion of (x + y)n where 'n' is a natural number. The expansion of (x + y)n has (n + 1) terms. This formula says:
We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
- General Term in (1 + x)n is nCr xr
- In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .