Greatest term in the expansion of (1 + t)^2n has greatest coefficient if and only if . The coefficient of x^5 in the expansion of (1 + x - 2x^2)^n is

JEE MainMathematicsToughBinomial Theorem

Greatest term in the expansion of (1 + t)^2n has greatest
coefficient if and only if
$t \in \left( \frac{9}{10},\frac{10}{9} \right)$ . The coefficient of
x^5 in the expansion of (1 + x - 2x^2)^n is

Show answer and solution

Greatest term in the expansion of (1 + x)^2n is having
greatest coefficient

\

If x

\in \left( \frac{n}{n + 1},\frac{n + 1}{n} \right) \Rightarrow

n = 9
coefficient of x^5 in (1 + x - 2x^2)^9

= \left( 1 + x + \left( - 2x^{2} \right) \right)^{9}

= \frac{9!(1)^{\alpha_{1}}(x)^{\alpha_{2}}\left( - 2x^{2} \right)^{\alpha_{3}}}{\alpha_{1}!\alpha_{2}!\alpha_{3}!}

------------------------------------
------------ ------------ ------------
6 1 2
4 5 0
------------------------------------
Coeff = 126

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

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