Binomial Theorem Question - None

Question

The coefficient of $x^5$ in the expansion of $(1+x)^{21} + (1+x)^{22} + \cdots + (1+x)^{30}$ is

Accepted Answer

Step 1: Recognize the sum as a geometric series:

$(1+x)^{21} + (1+x)^{22} + \cdots + (1+x)^{30} = (1+x)^{21}[1 + (1+x) + (1+x)^2 + \cdots + (1+x)^9]$

Step 2: Apply the geometric series formula with first term a = 1, ratio r = (1+x), and 10 terms:

$= (1+x)^{21} \cdot \frac{(1+x)^{10} - 1}{(1+x) - 1} = (1+x)^{21} \cdot \frac{(1+x)^{10} - 1}{x}$

Step 3: Rewrite as:

$= \frac{(1+x)^{31} - (1+x)^{21}}{x}$

Step 4: The coefficient of $x^5$ in the numerator requires the coefficient of $x^6$ in $(1+x)^{31}$ minus the coefficient of $x^6$ in $(1+x)^{21}$ (since dividing by x shifts the power down by 1):

$= \binom{31}{6} - \binom{21}{6}$

Step 5: Calculate: $\binom{31}{6} - \binom{21}{6} = 736457 - 54264 = 682193$

∴ Answer: C

Question:

Step-by-Step Solution

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