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Math  /  Algebra

Question(c) (n1)+2(n2)+3(n3)+⋯+n(nn)=n2n−1\binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}+\cdots+n\binom{n}{n}=n 2^{n-1}. [Hint: After expanding n(1+b)n−1n(1+b)^{n-1} by the binomial theorem, let b=1b=1; note also that n(n−1k)=(k+1)(nk+1)⋅]\left.n\binom{n-1}{k}=(k+1)\binom{n}{k+1} \cdot\right]

Studdy Solution

STEP 1

1. We are given a binomial identity involving binomial coefficients.
2. The goal is to prove that the given sum equals n⋅2n−1 n \cdot 2^{n-1} .
3. We will use the binomial theorem and properties of binomial coefficients to prove the identity.

STEP 2

1. Expand n(1+b)n−1 n(1+b)^{n-1} using the binomial theorem.
2. Substitute b=1 b = 1 into the expansion.
3. Use the property n(n−1k)=(k+1)(nk+1) n\binom{n-1}{k} = (k+1)\binom{n}{k+1} to relate terms.
4. Simplify the expression to show it equals n⋅2n−1 n \cdot 2^{n-1} .

STEP 3

Start by expanding n(1+b)n−1 n(1+b)^{n-1} using the binomial theorem:
The binomial theorem states that:
(1+b)n−1=∑k=0n−1(n−1k)bk (1+b)^{n-1} = \sum_{k=0}^{n-1} \binom{n-1}{k} b^k
Multiplying both sides by n n , we get:
n(1+b)n−1=n∑k=0n−1(n−1k)bk n(1+b)^{n-1} = n \sum_{k=0}^{n-1} \binom{n-1}{k} b^k

STEP 4

Substitute b=1 b = 1 into the expanded expression:
n(1+1)n−1=n∑k=0n−1(n−1k)⋅1k n(1+1)^{n-1} = n \sum_{k=0}^{n-1} \binom{n-1}{k} \cdot 1^k
This simplifies to:
n⋅2n−1=n∑k=0n−1(n−1k) n \cdot 2^{n-1} = n \sum_{k=0}^{n-1} \binom{n-1}{k}

STEP 5

Use the property n(n−1k)=(k+1)(nk+1) n\binom{n-1}{k} = (k+1)\binom{n}{k+1} to relate terms:
n∑k=0n−1(n−1k)=∑k=0n−1(k+1)(nk+1) n \sum_{k=0}^{n-1} \binom{n-1}{k} = \sum_{k=0}^{n-1} (k+1)\binom{n}{k+1}
Re-index the sum by letting j=k+1 j = k+1 , then k=j−1 k = j-1 , and the limits of summation change accordingly:
∑j=1nj(nj) \sum_{j=1}^{n} j \binom{n}{j}

STEP 6

Recognize that the expression ∑j=1nj(nj) \sum_{j=1}^{n} j \binom{n}{j} is equivalent to the original problem statement:
(n1)+2(n2)+3(n3)+⋯+n(nn) \binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \cdots + n\binom{n}{n}
Thus, we have shown:
(n1)+2(n2)+3(n3)+⋯+n(nn)=n⋅2n−1 \binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \cdots + n\binom{n}{n} = n \cdot 2^{n-1}
The identity is proven, and the expression equals n⋅2n−1 n \cdot 2^{n-1} .

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