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Question

If the coefficient of $$2$$nd, $$3$$rd and $$4$$th terms in the expansion of $${ \left( 1+x \right)  }^{ 2n }$$ are in A.P. Show that $$2{ n }^{ 2 }-9n+7=0$$


Solution

$$ (1+x)^{2n}$$
$$ 2nd term = (2n)c_{1}x$$
$$ 3rd term = (2n)c_{2}x^{2}$$
$$ 4th term = (2n)c_{3}x^{2}$$
coefficient in AP $$ \Rightarrow (2n)c_{1}+(2n)c^{3}=2\times (2n)c_{2}$$
$$\displaystyle \therefore 2n +\frac{(2n)(2n-1)(2n-2)}{3!}= 2\times \frac{2n(2n-1)}{2!} $$
$$ \displaystyle \therefore 2n(1+\frac{(2n-1)(2n-2)}{6})=2n(2n-1)$$
$$\displaystyle \therefore (6+4n^{2-6n+2})=12n-6$$
$$\displaystyle \therefore 4n^{2}-18n+14=0$$
$$\displaystyle  \therefore 2n^{2}-9n+7=0$$

Mathematics

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