Sequences & Series
Geometric Series
Grade 11
Question:
<p>The sum of
\[(x + 2)^{n-1} + (x + 2)^{n-2}(x + 1) + (x + 2)^{n-3}(x + 1)^2 + \ldots + (x + 1)^{n-1}\]
is equal to</p>
<p>(a) \((x + 2)^{n-2} - (x + 1)^n\)</p>
<p>(b) \(\left(\frac{(x+2)^n - (x+1)^n}{x+1}\right)\)</p>
<p>(c) \((x + 2)^n - (x + 1)^n\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Recognize this as a geometric series with ratio (x+1)/(x+2) and apply the geometric series sum formula.
<p>This is a geometric series with first term $a = (x+2)^{n-1}$ and common ratio $r = \frac{x+1}{x+2}$</p><p>Number of terms = n</p><p>Using the formula for sum of geometric series:</p><p>$$S = \frac{a(1-r^n)}{1-r} = \frac{(x+2)^{n-1}\left(1 - \left(\frac{x+1}{x+2}\right)^n\right)}{1 - \frac{x+1}{x+2}}$$</p><p>$$= \frac{(x+2)^{n-1}\left(1 - \frac{(x+1)^n}{(x+2)^n}\right)}{\frac{1}{x+2}}$$</p><p>$$= (x+2)^n\left(\frac{(x+2)^n - (x+1)^n}{(x+2)^n}\right)$$</p><p>$$= (x+2)^n - (x+1)^n$$</p><p>∴ Answer is (c)</p>
Correct Answer: C