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JEE Main 2018
Binomial Theorem
Binomial Theorem
Hard

Question

Statement - 1 : ∑r=0n(r+1) nCr=(n+2)2n−1.\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} Statement - 2 : ∑r=0n(r+1) nCrxr=(1+x)n+nx(1+x)n−1.\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.}

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Solution

This problem tests your understanding of the Binomial Theorem and its applications, particularly evaluating sums involving binomial coefficients and powers of xx. We will use various identities and techniques, including splitting sums, applying combinatorial identities, and differentiating binomial expansions.

Key Concepts and Techniques

  1. Binomial Expansion: The fundamental identity is: (1+x)n=∑r=0nnCrxr=nC0+nC1x+nC2x2+⋯+nCnxn(1+x)^n = \sum_{r=0}^n {}^n C_r x^r = {}^n C_0 + {}^n C_1 x + {}^n C_2 x^2 + \dots + {}^n C_n x^n

    • Tip: This is the starting point for many problems involving sums of binomial coefficients.

  2. Sum of Binomial Coefficients: By setting x=1x=1 in the binomial expansion: ∑r=0nnCr=nC0+nC1+⋯+nCn=(1+1)n=2n\sum_{r=0}^n {}^n C_r = {}^n C_0 + {}^n C_1 + \dots + {}^n C_n = (1+1)^n = 2^n

  3. Combinatorial Identity for r⋅nCrr \cdot {}^n C_r: This identity simplifies expressions where rr is multiplied by a binomial coefficient: r⋅nCr=n⋅n−1Cr−1r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}

    • Why it's useful: It transforms a term involving rr and nCr{}^n C_r into a simpler binomial coefficient n−1Cr−1{}^{n-1} C_{r-1} multiplied by nn, allowing us to sum it more easily.
    • Proof: r⋅n!r!(n−r)!=n!(r−1)!(n−r)!r \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r)!}. Also, n⋅(n−1)!(r−1)!((n−1)−(r−1))!=n⋅(n−1)!(r−1)!(n−r)!=n!(r−1)!(n−r)!n \cdot \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!} = n \cdot \frac{(n-1)!}{(r-1)!(n-r)!} = \frac{n!}{(r-1)!(n-r)!}.
    • Note: This identity is valid for r≥1r \ge 1. For r=0r=0, r⋅nCr=0r \cdot {}^n C_r = 0, so sums involving r⋅nCrr \cdot {}^n C_r can safely start from r=1r=1.

  4. Differentiation of Binomial Expansion: Differentiating the binomial expansion with respect to xx is a powerful technique for sums involving r⋅nCrxrr \cdot {}^n C_r x^r or r⋅nCrxr−1r \cdot {}^n C_r x^{r-1}.

    • Start with (1+x)n=∑r=0nnCrxr(1+x)^n = \sum_{r=0}^n {}^n C_r x^r.
    • Differentiate both sides with respect to xx: ddx((1+x)n)=ddx(∑r=0nnCrxr)\frac{d}{dx} \left( (1+x)^n \right) = \frac{d}{dx} \left( \sum_{r=0}^n {}^n C_r x^r \right) n(1+x)n−1=∑r=1nrnCrxr−1n(1+x)^{n-1} = \sum_{r=1}^n r {}^n C_r x^{r-1}
    • Why it's useful: The differentiation process brings down the exponent rr, creating the term r⋅nCrr \cdot {}^n C_r. Multiplying by xx (if needed) can adjust the power of xx to match the required sum. Note that the sum starts from r=1r=1 because the r=0r=0 term (nC0{}^n C_0) is a constant whose derivative is zero.

Verifying Statement - 1

Statement - 1: ∑r=0n(r+1) nCr=(n+2)2n−1.\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.}

Let's evaluate the sum on the left-hand side (LHS): LHS=∑r=0n(r+1) nCrLHS = \sum_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}}

Step 1: Decompose the sum. We can split the sum into two simpler parts, as (r+1)nCr=r⋅nCr+1⋅nCr(r+1){}^n C_r = r \cdot {}^n C_r + 1 \cdot {}^n C_r: LHS=∑r=0nr⋅nCr+∑r=0nnCrLHS = \sum_{r = 0}^n {r \cdot {}^n{C_r}} + \sum_{r = 0}^n {{}^n{C_r}}

Step 2: Evaluate the second part, ∑r=0nnCr\sum_{r = 0}^n {{}^n{C_r}}. This is the direct sum of all binomial coefficients for a given nn. Using the Sum of Binomial Coefficients identity: ∑r=0nnCr=2n\sum_{r = 0}^n {{}^n{C_r}} = 2^n

Step 3: Evaluate the first part, ∑r=0nr⋅nCr\sum_{r = 0}^n {r \cdot {}^n{C_r}}.

  • First, notice that for r=0r=0, the term r⋅nCr=0⋅nC0=0r \cdot {}^n C_r = 0 \cdot {}^n C_0 = 0. So, we can rewrite the sum starting from r=1r=1: ∑r=0nr⋅nCr=∑r=1nr⋅nCr\sum_{r = 0}^n {r \cdot {}^n{C_r}} = \sum_{r = 1}^n {r \cdot {}^n{C_r}}
  • Now, apply the Combinatorial Identity r⋅nCr=n⋅n−1Cr−1r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}. This simplifies the term r⋅nCrr \cdot {}^n C_r: ∑r=1nn⋅n−1Cr−1\sum_{r = 1}^n {n \cdot {}^{n - 1}{C_{r - 1}}}
  • Factor out nn, as it is a constant with respect to the summation variable rr: n∑r=1nn−1Cr−1n \sum_{r = 1}^n {{}^{n - 1}{C_{r - 1}}}
  • To evaluate this sum, let's change the index of summation. Let k=r−1k = r-1.
    • When r=1r=1, k=0k=0.
    • When r=nr=n, k=n−1k=n-1. The sum transforms into: n∑k=0n−1n−1Ckn \sum_{k = 0}^{n - 1} {{}^{n - 1}{C_k}}
  • This is again the sum of all binomial coefficients, but for (n−1)(n-1). Using the Sum of Binomial Coefficients identity: n⋅2n−1n \cdot 2^{n - 1}

Step 4: Combine the results from Step 2 and Step 3. LHS=(n⋅2n−1)+(2n)LHS = \left( n \cdot 2^{n - 1} \right) + \left( 2^n \right) To simplify, factor out 2n−12^{n-1} from both terms. Remember 2n=2⋅2n−12^n = 2 \cdot 2^{n-1}: LHS=2n−1(n+2)LHS = 2^{n - 1} \left( n + 2 \right) This matches the right-hand side (RHS) of Statement - 1. Therefore, Statement - 1 is True.

Verifying Statement - 2

Statement - 2: ∑r=0n(r+1) nCrxr=(1+x)n+nx(1+x)n−1.\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.}

Let's evaluate the sum on the left-hand side (LHS): LHS=∑r=0n(r+1) nCrxrLHS = \sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r}}

Step 1: Decompose the sum. Similar to Statement 1, we split the sum into two parts: LHS=∑r=0nr⋅nCrxr+∑r=0nnCrxrLHS = \sum\limits_{r = 0}^n {r \cdot {}^n{C_r}{x^r}} + \sum\limits_{r = 0}^n {{}^n{C_r}{x^r}}

Step 2: Evaluate the second part, ∑r=0nnCrxr\sum_{r = 0}^n {{}^n{C_r}{x^r}}. This is the direct application of the Binomial Expansion for (1+x)n(1+x)^n: ∑r=0nnCrxr=(1+x)n\sum_{r = 0}^n {{}^n{C_r}{x^r}} = (1+x)^n

Step 3: Evaluate the first part, ∑r=0nr⋅nCrxr\sum_{r = 0}^n {r \cdot {}^n{C_r}{x^r}}. This sum can be effectively evaluated using the Differentiation of Binomial Expansion technique.

  • Start with the binomial expansion: (1+x)n=∑r=0nnCrxr(1+x)^n = \sum_{r=0}^n {{}^n{C_r}{x^r}}
  • Differentiate both sides with respect to xx. We do this because differentiating xrx^r brings down rr as a coefficient, which is what we need in our sum: ddx((1+x)n)=ddx(∑r=0nnCrxr)\frac{d}{dx} \left( (1+x)^n \right) = \frac{d}{dx} \left( \sum_{r=0}^n {{}^n{C_r}{x^r}} \right) n(1+x)n−1=∑r=1nr⋅nCrxr−1n(1+x)^{n-1} = \sum_{r=1}^n {r \cdot {}^n{C_r}{x^{r-1}}} (Note: The term for r=0r=0 on the RHS, nC0x0{}^n C_0 x^0, is a constant, so its derivative is 0. Thus, the sum starts from r=1r=1.)
  • Now, we need r⋅nCrxrr \cdot {}^n C_r x^r, not r⋅nCrxr−1r \cdot {}^n C_r x^{r-1}. To achieve this, multiply both sides of the equation by xx: x⋅n(1+x)n−1=x⋅∑r=1nr⋅nCrxr−1x \cdot n(1+x)^{n-1} = x \cdot \sum_{r=1}^n {r \cdot {}^n{C_r}{x^{r-1}}} nx(1+x)n−1=∑r=1nr⋅nCrxrnx(1+x)^{n-1} = \sum_{r=1}^n {r \cdot {}^n{C_r}{x^r}}
  • The sum ∑r=1nr⋅nCrxr\sum_{r=1}^n {r \cdot {}^n{C_r}{x^r}} is equivalent to ∑r=0nr⋅nCrxr\sum_{r=0}^n {r \cdot {}^n{C_r}{x^r}} because the r=0r=0 term (0⋅nC0x00 \cdot {}^n C_0 x^0) is zero. So, we have: ∑r=0nr⋅nCrxr=nx(1+x)n−1\sum_{r = 0}^n {r \cdot {}^n{C_r}{x^r}} = nx{{\left( {1 + x} \right)}^{n - 1}}

Step 4: Combine the results from Step 2 and Step 3. LHS=(nx(1+x)n−1)+((1+x)n)LHS = \left( nx{{\left( {1 + x} \right)}^{n - 1}} \right) + \left( {(1+x)^n} \right) LHS=(1+x)n+nx(1+x)n−1LHS = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}} This matches the right-hand side (RHS) of Statement - 2. Therefore, Statement - 2 is True.

Relationship between Statement - 1 and Statement - 2

We have found that both Statement - 1 and Statement - 2 are true. Now we need to determine if Statement - 2 is a correct explanation for Statement - 1.

Statement - 2 provides a general identity involving xx: ∑r=0n(r+1) nCrxr=(1+x)n+nx(1+x)n−1.\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.}

To see if this explains Statement - 1, let's substitute a specific value for xx into Statement - 2. Notice that Statement - 1 is a sum of terms without xrx^r. This suggests setting x=1x=1 in Statement - 2.

Substitute x=1x=1 into the identity from Statement - 2: ∑r=0n(r+1) nCr(1)r=(1+1)n+n(1)(1+1)n−1\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{(1)^r}} = {{\left( {1 + 1} \right)}^n} + n(1){{\left( {1 + 1} \right)}^{n - 1}} ∑r=0n(r+1) nCr=2n+n2n−1\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}} = {2^n} + n{2^{n - 1}} ∑r=0n(r+1) nCr=2n−1(2+n)\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}} = {2^{n - 1}} \left( 2 + n \right) ∑r=0n(r+1) nCr=(n+2)2n−1\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}} = \left( {n + 2} \right){2^{n - 1}}

This result exactly matches Statement - 1. Therefore, Statement - 2 is a more general identity, and Statement - 1 is a special case derived from Statement - 2 by setting x=1x=1. This means Statement - 2 is indeed a correct explanation for Statement - 1.

Conclusion

Both Statement - 1 and Statement - 2 are true, and Statement - 2 provides a general formula from which Statement - 1 can be directly derived by setting x=1x=1.

The final answer is B\boxed{B}.

Key Takeaways:

  • Splitting sums: A common strategy when the general term has multiple components (e.g., (r+1)nCr(r+1){}^n C_r).
  • Combinatorial identity r⋅nCr=n⋅n−1Cr−1r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}: Essential for simplifying sums involving rr without xx.
  • Differentiation technique: The most elegant method for evaluating sums of the form ∑r⋅nCrxr\sum r \cdot {}^n C_r x^r or ∑r⋅nCrxr−1\sum r \cdot {}^n C_r x^{r-1}. Remember to adjust the power of xx by multiplying by xx if needed.
  • General vs. Specific: Often, a more general identity (like Statement - 2) can explain a specific numerical identity (like Statement - 1) by substituting a particular value.
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