The coefficient of x^r (0 le r le n - 1) in t | vedclass

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Question

(B) The given expression is a geometric series with $n$ terms,where the first term $a = (x + 2)^{n-1}$ and the common ratio $q = \frac{x+1}{x+2}$.
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Vedclass · Accepted Answer

$^nC_r (2^{n-r} - 1)$

Vedclass · Answer

(B) The given expression is a geometric series with $n$ terms,where the first term $a = (x + 2)^{n-1}$ and the common ratio $q = \frac{x+1}{x+2}$.
Using the sum formula $S_n = a \frac{1 - q^n}{1 - q}$:
$E = (x + 2)^{n-1} \left[ \frac{1 - (\frac{x+1}{x+2})^n}{1 - \frac{x+1}{x+2}} \right]$
$E = (x + 2)^{n-1} \left[ \frac{\frac{(x+2)^n - (x+1)^n}{(x+2)^n}}{\frac{x+2 - x - 1}{x+2}} \right]$
$E = (x + 2)^{n-1} \left[ \frac{(x+2)^n - (x+1)^n}{(x+2)^n} \cdot (x+2) \right]$
$E = (x + 2)^n - (x + 1)^n$
Expanding using the binomial theorem:
$(2 + x)^n - (1 + x)^n = \sum_{r=0}^{n} {^nC_r} 2^{n-r} x^r - \sum_{r=0}^{n} {^nC_r} 1^{n-r} x^r$
$= \sum_{r=0}^{n} {^nC_r} (2^{n-r} - 1) x^r$
Thus,the coefficient of $x^r$ is $^nC_r (2^{n-r} - 1)$.

The coefficient of $x^r$ $(0 \le r \le n - 1)$ in the expression: $(x + 2)^{n-1} + (x + 2)^{n-2}(x + 1) + (x + 2)^{n-3}(x + 1)^2 + \dots + (x + 1)^{n-1}$ is:

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