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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
The sum of all possible values of $n \in \mathbb{N}$, so that the coefficients of $x$, $x^2$ and $x^3$ in the expansion of $(1 + x)^2(1 + x)^n$, are in arithmetic progression is:
Solution
$(x^4 + 2x^2 + 1)\left({}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + {}^nC_3 x^3 + \cdots\right)$
Coefficient of $x$ = ${}^nC_1$
Coefficient of $x^2$ = $2 + {}^nC_2$
Coefficient of $x^3$ = $2\cdot {}^nC_1 + {}^nC_3$
$= 2n + \frac{n(n-1)(n-2)}{6}$
Now according to question
$n + 2n + \frac{n(n-1)(n-2)}{6} = 2\left(2 + \frac{n(n-1)}{2}\right)$
$3n + \frac{n(n-1)(n-2)}{6} = 4 + n(n-1)$
$\Rightarrow n^3 - 9n^2 + 26n - 24 = 0$
$\Rightarrow n = 2, 3, 4$
Now checking for $n = 2$
Coeff of $x = 2$, coeff of $x^2 = 3$, coeff of $x^3 = 4$ ⇒ are in A.P.
Required sum $= 2 + 3 + 4 = 9$
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