Binomial Theorem
Sum of coefficients in binomial expansion
Grade 11
Question:
<p>If \((1-x)^{-n} = a_0 + a_1 x + a_2 x^2 + \cdots + a_r x^r + \cdots\), then \(a_0 + a_1 + a_2 + \cdots + a_r\) is equal to</p>
<p>(1) \(\dfrac{n(n+1)(n+2)\cdots(n+r)}{r!}\)</p>
<p>(2) \(\dfrac{(n+1)(n+2)\cdots(n+r)}{r!}\)</p>
<p>(3) \(\dfrac{n(n+1)(n+2)\cdots(n+r-1)}{r!}\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: To find the sum of coefficients up to the r-th term, substitute x=1 in the expansion, but recognize that the infinite series $(1-x)^{-n}$ at x=1 requires careful handling of the partial sum up to $x^r$.
<p><strong>Step 1:</strong> Expand $(1-x)^{-n}$ using the binomial series formula:</p><p>$(1-x)^{-n} = \sum_{k=0}^{\infty} \binom{n+k-1}{k} x^k$</p><p>Therefore: $a_r = \binom{n+r-1}{r}$</p><p><strong>Step 2:</strong> Find the sum $S = a_0 + a_1 + a_2 + \cdots + a_r$:</p><p>$S = \sum_{k=0}^{r} \binom{n+k-1}{k}$</p><p><strong>Step 3:</strong> Apply the hockey-stick identity: $\sum_{k=0}^{r} \binom{n+k-1}{k} = \binom{n+r}{r}$</p><p><strong>Step 4:</strong> Alternatively, use the partial derivative approach: differentiating $(1-x)^{-n}$ successively and evaluating at x=0 gives the coefficient pattern, and summing from 0 to r yields the identity.</p><p>∴ Answer: $\binom{n+r}{r}$ or equivalently $\binom{n+r}{n}$ (Option A)</p>
Correct Answer: A