You are not logged in.
1. Need to Use Pascal's Formula and induction on n to prove the Binomial Theorem:
(DONT UNDERSTAND THE PART IN BOLD) Correct otherwise??
Base Case: (a+b)^0 = 1 = \sum_{k=0}^0 { 0 \choose k } a^{0-k}b^k.
For the inductive step, assume the theorem holds when the exponent is m. Then for n = m + 1
(a+b)^{m+1} = a(a+b)^m + b(a+b)^m \,
= a \sum_{k=0}^m { m \choose k } a^{m-k} b^k + b \sum_{j=0}^m { m \choose j } a^{m-j} b^j by the inductive hypothesis
= \sum_{k=0}^m { m \choose k } a^{m-k+1} b^k + \sum_{j=0}^m { m \choose j } a^{m-j} b^{j+1} by multiplying through by a and b
= a^{m+1} + \sum_{k=1}^m { m \choose k } a^{m-k+1} b^k + \sum_{j=0}^m { m \choose j } a^{m-j} b^{j+1} by pulling out the k = 0 term
= a^{m+1} + \sum_{k=1}^m { m \choose k } a^{m-k+1} b^k + \sum_{k=1}^{m+1} { m \choose k-1 }a^{m-k+1}b^{k} by letting j = k − 1
= a^{m+1} + \sum_{k=1}^m { m \choose k } a^{m-k+1}b^k + \sum_{k=1}^{m} { m \choose k-1 }a^{m+1-k}b^{k} + b^{m+1} by pulling out the k = m + 1 term from the RHS
= a^{m+1} + b^{m+1} + \sum_{k=1}^m \left[ { m \choose k } + { m \choose k-1 } \right] a^{m+1-k}b^k by combining the sums
= a^{m+1} + b^{m+1} + \sum_{k=1}^m { m+1 \choose k } a^{m+1-k}b^k from Pascal's rule
= \sum_{k=0}^{m+1} { m+1 \choose k } a^{m+1-k}b^k by adding in the m + 1 terms.
2. Summing the cubes?
a)Prove directly that m^3=6(m choose 3)+6(m choose 2)+m
b)Use part (a) to prove that the sum(from i=1 to n) i^3 = [(n(n+1)/2]^2.(without using induction)
c)Prove part (a) by counting a set in two ways (hint: Count the ordered triples that can be formed from [m].)
Really just not sure where to go here... maybe some hint may help? I think I can do it... not sure where to start?
Last edited by clooneyisagenius (2007-04-08 14:48:44)
Offline
sorry, i dont know how to make equations?!
Offline
see the sticky
Numbers are the essence of the Universe
Offline
1. Need to Use Pascal's Formula and induction on n to prove the Binomial Theorem:
(DONT UNDERSTAND THE PART IN BOLD) Correct otherwise??Base Case:
.
For the inductive step, assume the theorem holds when the exponent is m. Then for n = m + 1
by the inductive hypothesis
by multiplying through by a and b
by pulling out the k = 0 term
by letting j = k − 1
by pulling out the k = m + 1 term from the RHS
by combining the sums
from Pascal's rule
by adding in the m + 1 terms.
To make equations, you need to enclose your LaTeX formulas in [math] [/math] tags.
The proof looks fine.
Offline