1. Weighted Binomial Sums
When dealing with sums like $\sum k \binom{k}{m}$, we use special identities. The key is $k \binom{k}{m} = (m+1)\binom{k+1}{m+1} – \binom{k+1}{m+2}$. This transforms products into differences of binomials with shifted indices, enabling application of the hockey-stick identity. Such weighted sums appear in combinatorics, probability theory, and discrete mathematics. Understanding these manipulations is crucial for JEE advanced problems. The technique involves pattern recognition, applying identities, and systematic simplification. Students should practice deriving these from first principles to build intuition about when and how to apply them in different contexts.
2. Hockey-Stick Identity
The hockey-stick identity states: $\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}$. Named for the hockey-stick shape formed in Pascal’s triangle when highlighting summed elements. It can be proved by induction or combinatorially: choosing $r+1$ objects from $n+1$ by considering the largest object chosen. If largest is $(i+1)$-th, choose $r$ from first $i$ objects, giving $\binom{i}{r}$ ways. Summing over all possible largest elements gives the identity. This is extremely useful for simplifying binomial sums and appears in numerous JEE problems involving series, combinatorics, and probability distributions.
3. Generating Functions
An alternative approach uses generating functions. The sum $\sum_{k=1}^{100} k(1+x)^k$ is a weighted geometric series. We know $(1+x) + (1+x)^2 + \ldots + (1+x)^n = \frac{(1+x)[(1+x)^n – 1]}{x}$. For weighted sums, differentiate or multiply by index. If $f(y) = \sum_{k=1}^{n} y^k = \frac{y(1-y^n)}{1-y}$, then $yf'(y) = \sum_{k=1}^{n} ky^k$. Setting $y=(1+x)$ and extracting coefficient of $x^{48}$ requires careful expansion. This method provides powerful tools for complex summations, particularly useful when direct combinatorial arguments become unwieldy.
4. Pascal’s Identity
Pascal’s identity, $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$, is fundamental to binomial manipulation. It reflects recursive structure of Pascal’s triangle, where each entry equals sum of two entries above it. In this problem, we use variations to relate binomials with different indices, allowing transformation of complicated expressions into forms matching answer choices. The identity has beautiful combinatorial interpretation: to choose $r$ from $n$ objects, either first object is included (choose $r-1$ from remaining $n-1$), or not (choose $r$ from remaining $n-1$). Mastering these manipulations requires practice with different index shifts and recognizing when to apply forward or backward forms efficiently.
Q
Why start from $k=48$ for $x^{48}$?
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In $(1+x)^k = \sum_{r=0}^{k} \binom{k}{r} x^r$, term $x^{48}$ appears only when $r=48$. Since $\binom{k}{48} = 0$ for $k < 48$, only $k \geq 48$ contribute.
Q
How derive identity for $k \binom{k}{m}$?
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Start with $k \binom{k}{m} = k \cdot \frac{k!}{m!(k-m)!}$. Express as $\frac{(k+1)!}{m!(k-m)!} – \frac{k!}{m!(k-m)!}$. First term equals $(m+1)\binom{k+1}{m+1}$, second gives $-\binom{k+1}{m+2}$ using Pascal’s identity.
Q
What is hockey-stick identity?
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Hockey-stick: $\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}$. Called hockey-stick because terms form this shape in Pascal’s triangle, with sum at bent end.
Q
Can use generating functions?
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Yes. Consider $\sum_{k=1}^{100} k(1+x)^k = (1+x) \frac{d}{d(1+x)} \sum_{k=1}^{100} (1+x)^k$. This gives geometric series to differentiate. Extract coefficient of $x^{48}$, though algebra is more involved.
Q
Which option is correct?
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Option B: $100 \cdot {}^{101}C_{49} – {}^{101}C_{50}$ is correct. After applying hockey-stick and simplifying, expression reduces to this form. Indices 101, 49, 50 arise naturally from upper limit 100 and power 48.
Q
How verify for small values?
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Test $n=3$, coefficient of $x^1$ in $(1+x) + 2(1+x)^2 + 3(1+x)^3$. Direct: $1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 = 14$. Formula: $3 \cdot \binom{4}{2} – \binom{4}{3} = 18 – 4 = 14$ ✓
Q
What if sum to infinity?
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For infinite sums $\sum_{k=1}^{\infty} k(1+x)^k$ with $|1+x| < 1$, use $\sum_{k=1}^{\infty} kr^k = \frac{r}{(1-r)^2}$. Extracting specific coefficients requires contour integration or advanced techniques.
Q
Why use $\binom{k+1}{m+1}$ not $\binom{k}{m+1}$?
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Index shift from $k$ to $k+1$ necessary for hockey-stick identity. When summing $\binom{k+1}{m+1}$ from $k=m$ to $n$, substituting $j=k+1$ gives $\sum_{j=m+1}^{n+1} \binom{j}{m+1} = \binom{n+2}{m+2}$.
Q
General formula for $x^m$ coefficient?
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Coefficient of $x^m$ is $\sum_{k=m}^{n} k\binom{k}{m} = n \cdot \binom{n+1}{m+1} – \binom{n+1}{m+2}$ when $m \leq n$. This generalizes our case with $n=100$, $m=48$.
Q
Similar problems in JEE?
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Yes, JEE frequently tests binomial coefficient summations. Problems with $\sum k^p (1+x)^k$ or $\sum \binom{n}{k} f(k)$ appear regularly. Key is recognizing which identity: hockey-stick, Vandermonde, or generating functions.