Question ID: #795
The sum of all possible values of $n\in N$ so that the coefficients of $x$, $x^{2}$ and $x^{3}$ in the expansion of $(1+x^{2})^{2}(1+x)^{n}$ are in arithmetic progression is:
- (1) 3
- (2) 7
- (3) 12
- (4) 9
Solution:
Let the expansion be $E = (1 + 2x^2 + x^4) \sum_{r=0}^{n} \, ^nC_r x^r$.
$$ E = (1 + 2x^2 + x^4)(^nC_0 + ^nC_1 x + ^nC_2 x^2 + ^nC_3 x^3 + \dots) $$
We need coefficients of $x, x^2, x^3$.
**Coefficient of $x$ ($A$):**
Only from $1 \cdot (^nC_1 x)$.
$$ A = ^nC_1 = n $$
**Coefficient of $x^2$ ($B$):**
From $1 \cdot (^nC_2 x^2)$ and $2x^2 \cdot (^nC_0)$.
$$ B = ^nC_2 + 2(^nC_0) = \frac{n(n-1)}{2} + 2 $$
**Coefficient of $x^3$ ($C$):**
From $1 \cdot (^nC_3 x^3)$ and $2x^2 \cdot (^nC_1 x)$.
$$ C = ^nC_3 + 2(^nC_1) = \frac{n(n-1)(n-2)}{6} + 2n $$
Given $A, B, C$ are in A.P., so $2B = A + C$.
$$ 2\left(\frac{n^2-n}{2} + 2\right) = n + \frac{n(n-1)(n-2)}{6} + 2n $$
$$ n^2 – n + 4 = 3n + \frac{n(n^2-3n+2)}{6} $$
Multiply by 6:
$$ 6n^2 – 6n + 24 = 18n + n^3 – 3n^2 + 2n $$
$$ 6n^2 – 6n + 24 = n^3 – 3n^2 + 20n $$
$$ n^3 – 9n^2 + 26n – 24 = 0 $$
By trial, put $n=2$: $8 – 36 + 52 – 24 = 60 – 60 = 0$. So $n=2$ is a root.
Dividing $(n-2)$ out from $n^3 – 9n^2 + 26n – 24$:
$$ n^2(n-2) – 7n(n-2) + 12(n-2) = 0 $$
$$ (n-2)(n^2 – 7n + 12) = 0 $$
$$ (n-2)(n-3)(n-4) = 0 $$
Possible values for $n$ are $2, 3, 4$.
Sum of values = $2 + 3 + 4 = 9$.
Ans. (4)
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