Let ${(1 + x^{2})}^{2} {(1 + x)}^{n} = A_{0} + A_{1} x + A_{2} X^{2} + \cdot \cdot \cdot$ If $A_{0}, A_{1}, A_{2}$ are in A.P. then the value of n is

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Solution

The correct options are
A 3
C 2
$(1 + 2 x^{2} + x^{4}) {(1 + x)}^{n} = A_{0} + A_{1} x + A_{2} x^{2} + . . . . . .$
Here $A_{0} = 1, A_{1} = n, A_{2} = 2 +^{n} C_{2}$
Given $A_{0}, A_{1}, A_{2}$ are in A.P.
$∴ n - 1 = 2 + \frac{n (n - 1)}{2} - n$
$\Rightarrow n^{2} - 5 n + 6 = 0 \Rightarrow n = 2, 3$

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Let (1+x2)2(1+x)n=A0+A1x+A2X2+⋅⋅⋅ If A0,A1,A2 are in A.P. then the value of n is

The correct options are A 3 C 2(1+2x2+x4)(1+x)n=A0+A1x+A2x2+...... Here A0=1,A1=n,A2=2+nC2Given A0,A1,A2 are in A.P.∴n−1=2+n(n−1)2−n⇒n2−5n+6=0⇒n=2,3

Let ${(1 + x^{2})}^{2} {(1 + x)}^{n} = A_{0} + A_{1} x + A_{2} X^{2} + \cdot \cdot \cdot$ If $A_{0}, A_{1}, A_{2}$ are in A.P. then the value of n is