use the binomial theorem to show that 0=(k=0) to n ((-1)^k)C(n,k) Solution Assum

Question

use the binomial theorem to show that 0=(k=0) to n ((-1)^k)C(n,k)

Explanation / Answer

Assume the Binomal Theorem work for some exponent n. (integer n>0) Then: (a + b)^(n + 1) = (a + b)(a + b)^n = (a + b) S[C(n,k) * a^k * b^(n-k) ] ... summing from k=0 to n = a*S[C(n,k) * a^k * b^(n-k) ] + b*S[C(n,k) * a^k * b^(n-k) ] = S[C(n,k) * a^*(k+1) * b^(n-k) ] + S[C(n,k) * a^k * b^(n+1-k) ] Rearrange these sums to combine like powers of a. To do that, note that the powers of a in the left sum go from 1 to n+1 and in the right sum they go from 0 to n. The common range is 1 to n, so rewrite each sum using S' to mean summing from i=1 to n, (instead of 0 to n), where i is the power of a in each term of the S' sum. The left sum has powers running from a^1 to a^(n+1), so (k) in that sum corresponds to a term with i=k+1, and there will be a leftover a^(n+1) term that's not in the S' sum: S[C(n,k) * a^(k+1) * b^(n-k) ] = S'[C(n,i-1) * a^i * b^(n-(i-1)) ] + C(n,n)*a^(n+1)*b^0 = S'[C(n,i-1) * a^i * b^(n+1-i) ] + C(n,n)*a^(n+1)*b^0 That, I think is the hardest step, relabeling k to (i-1) to match powers of a. The right sum has powers of a running from 0 to n, so there's a leftover a^0 term, and any other term k has a power a^k, so k corresponds to i (the power of a) in this sum: S[C(n,k) * a^k * b^(n+1-k) ] = C(n,0)*a^0*b^(n+1) + S'[C(n,i) * a^i * b^(n+1-i) ] Adding the rewritten sums together gives another way to write (a + b)^(n + 1). Each S' has the same index variable i and range 1 to n, and the same powers a^i*b^(n+1-i), so you can just add the coefficients in the summands. So: (a + b)^(n+1) = C(n,0)*a^0*b^(n+1) + S'[ (C(n,i-1) + C(n,i))*a^i*b^(n+1-i) ] + C(n,n)a^(n+1)b^0 For any 0

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.