If n is a rational number and x is a real number such that |x|is less than 1, then
(1+x) n = 1 = nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3!+…+n(n-1)(n-2)…(n-r+1)xr/r!+...+ ∞
Remarks
The condition |x|<1 is unnecessary when n is a whole number.
When n is not a whole number, then the condition |x|<1 is necessary.
The terms are infinite when n is not an whole number. When is it an whole number the series become finite as one of the terms will become zero in the coefficient at some point in time.
When n is a positive integer, there will be n+1 terms
To expand (x+a)n proceed as follows:
(x+a)n = {a(1 + x/a)} n= an(1 + x/a) n; (substitute x’ = x/a and proceed)
Showing posts with label Binomial theorem. Show all posts
Showing posts with label Binomial theorem. Show all posts
Greatest Term in the expansion of (x+a)n
(i) Write (r+1)th term = T(r+1) and rth term = T (r)from the given expression.
(ii) Find T(r+1)/T(r)
(iii) Put T(r+1)/T(r)>1
(iv) Solve the inequality for r to get an inequality of the form rm
If m is an integer, the mth and (m+1)th terms are equal in magnitude and these two are the greatest terms.
If m is not an integer, then obtain integral part of m, say, k. In this case (k+1) term is the greatest term
(ii) Find T(r+1)/T(r)
(iii) Put T(r+1)/T(r)>1
(iv) Solve the inequality for r to get an inequality of the form rm
If m is an integer, the mth and (m+1)th terms are equal in magnitude and these two are the greatest terms.
If m is not an integer, then obtain integral part of m, say, k. In this case (k+1) term is the greatest term
Middle Terms in Bi' expression
Binomial expression (1+x)n contains (n+1) terms when n is a natural number.
If n is even the ((n/2) +1) th term is middle term.
If n is odd then ((n+1)/2) th and ((n+3)/2)th terms are two middle terms
If n is even the ((n/2) +1) th term is middle term.
If n is odd then ((n+1)/2) th and ((n+3)/2)th terms are two middle terms
Important from Binomial Theorem
1. total number of terms in the expansion = n+1
2. The sum of indices of x and a in each term is n.
3. the coefficients of terms equidistant from the beginning and the end are equal.
4. (x-a)n = (r = 0 to n)Σ ((-1) r*nCrxn-rar
The terms in the expansion of(x-a)n are alternatively positive and negative, the last term is positive or negative according as n is even or odd.
5. (1+x) n = (r = 0 to n)Σ nCrxr
you get it by putting x =1 and a = x in the expression for (x+a)n.
6. (x+1) n = (r = 0 to n)Σ nCrxn-r
7. (1-x) n = (r = 0 to n)Σ(-1)r* nCrxr
8. (x+a) n +(x-a) n = 2[nC0xna0 + nC2xn-2a2 + nC4xn-4a4+ ...]
9. General term in a binomial expansion (r+1) term in (x+a) n= nCrxn-rar
2. The sum of indices of x and a in each term is n.
3. the coefficients of terms equidistant from the beginning and the end are equal.
4. (x-a)n = (r = 0 to n)Σ ((-1) r*nCrxn-rar
The terms in the expansion of(x-a)n are alternatively positive and negative, the last term is positive or negative according as n is even or odd.
5. (1+x) n = (r = 0 to n)Σ nCrxr
you get it by putting x =1 and a = x in the expression for (x+a)n.
6. (x+1) n = (r = 0 to n)Σ nCrxn-r
7. (1-x) n = (r = 0 to n)Σ(-1)r* nCrxr
8. (x+a) n +(x-a) n = 2[nC0xna0 + nC2xn-2a2 + nC4xn-4a4+ ...]
9. General term in a binomial expansion (r+1) term in (x+a) n= nCrxn-rar
Binomial theorem
If x and a are real numbers, then for all n Є N
(x+a)n
= nC0xna0 + nC1xn-1a1 +nC2xn-2a2 + ...+nCrxn-rar+ ...+nCn-1x1an-1+nCnx0an
(x+a)n
= (r = 0 to n)Σ nCrxn-rar
(x+a)n
= nC0xna0 + nC1xn-1a1 +nC2xn-2a2 + ...+nCrxn-rar+ ...+nCn-1x1an-1+nCnx0an
(x+a)n
= (r = 0 to n)Σ nCrxn-rar
Pascal’s Triangle of Binomial coefficient
Expansions of (x+a)n as n = 0 to n
Each row is bounded by 1 on both sides
--- (The first row has only one item 1)First row is for n = 0
--- (The second row has two items 1 1)Second row is for n = 1
Any entry except the first and last entry in a row is the sum of two entries in the preceding row, one on the immediate left and the other on the immediate right.
So the third row (n = 2) is 1 2 1 (The second row elements are 1 1).
Each row is bounded by 1 on both sides
--- (The first row has only one item 1)First row is for n = 0
--- (The second row has two items 1 1)Second row is for n = 1
Any entry except the first and last entry in a row is the sum of two entries in the preceding row, one on the immediate left and the other on the immediate right.
So the third row (n = 2) is 1 2 1 (The second row elements are 1 1).
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