If (1 - x + 2x^2)^n = a_0 + a_1x + a_2x^2 + d | vedclass

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(D) Given: $(1 - x + 2x^2)^n = a_0 + a_1x + a_2x^2 + \dots$ .....$(1)$Putting $x = 0$,we get $a_0 = 1$.Differentiating equation $(1)$ with respect to

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no value of $n$

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(D) Given: $(1 - x + 2x^2)^n = a_0 + a_1x + a_2x^2 + \dots$ .....$(1)$Putting $x = 0$,we get $a_0 = 1$.Differentiating equation $(1)$ with respect to $x$:$n(1 - x + 2x^2)^{n-1}(-1 + 4x) = a_1 + 2a_2x + \dots$ .....$(2)$Putting $x = 0$,we get $a_1 = -n$.Differentiating equation $(2)$ with respect to $x$ again:$n(n-1)(1 - x + 2x^2)^{n-2}(-1 + 4x)^2 + n(1 - x + 2x^2)^{n-1}(4) = 2a_2 + \dots$ .....$(3)$Putting $x = 0$ in equation $(3)$:$n(n-1)(1) + 4n = 2a_2$$n^2 - n + 4n = 2a_2 \Rightarrow 2a_2 = n^2 + 3n$.Since $a_0, a_1, a_2$ are in $A.P.$,the condition is $2a_1 = a_0 + a_2$ or $2a_2 = a_0 + a_1$ depending on the order. Assuming the standard order $a_0, a_1, a_2$ are in $A.P.$,then $2a_1 = a_0 + a_2$. However,if the problem implies $a_0, a_1, a_2$ are in $A.P.$,then $2a_1 = a_0 + a_2$. Let's test $2a_1 = a_0 + a_2$: $2(-n) = 1 + \frac{n^2+3n}{2} \Rightarrow -4n = 2 + n^2 + 3n \Rightarrow n^2 + 7n + 2 = 0$. Roots are not natural numbers.If the order is $a_0, a_2, a_1$ as per the provided solution logic,$2a_2 = a_0 + a_1 \Rightarrow n^2 + 3n = 1 - n \Rightarrow n^2 + 4n - 1 = 0$. Roots are $-2 \pm \sqrt{5}$,which are not natural numbers. Thus,no value of $n$ exists.

If $(1 - x + 2x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,where $n \in N$,$x \in R$,and $a_0, a_1, a_2$ are in Arithmetic Progression $(A.P.)$,then there exists:

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