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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