Can someone teach me binomial theorem for natural numbers and for any index
#binomial theorem
meager swift
night boneBOT
Can someone teach me binomial theorem for natural numbers and for any index
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lean spade
Naturally, the binomial theorem is used for expanding an algebraic expression of two terms raised to the power of a natural number, but one can also use it to expand expressions involving more than two terms
To understand what it says, one first learns what a binomial coefficient is
meager swift
Pascal triangle
lean spade
Pascal triangle
Right, using the Pascal Triangle is an excellent way to derive the binomial coefficients without the help of a calculator
Factorial of a natural number n, written as n!, is defined as follows
n!=n×(n-1)×(n-2)...×3×2×1
So it is the product of all the natural numbers starting from n to 1
lean spade
Why do we use combination in binomial expansions
Can you explain what do you mean by combinations
lean spade
nCr
It is a short way to write the coefficient of the (r-1)th term
Suppose we are considering the coefficient of the first term, then the coefficient is nC0, which is always 1
meager swift
Yes
lean spade
Then the coefficient of the second term, nC1, which is always n
meager swift
Okay
lean spade
Is there anything in particular which you don't understand?
You can always to use pascal triangle to compute the binomial coefficients, just in case the power is too large (suppose 10) and you don't have time to draw the triangle, you can just compute it by a calculator
lean spade
What about non natural index
OK, is it ok for now if we raise the expression to a rational number?
meager swift
Yes
lean spade
The binomial theorem tells us how to do things for natural number
The expression would be 1 if we consider power 0
Now we deal with negative integers
A negative integer n' can always be written as -n, where n is natural
meager swift
Okay
vale crescentBOT
ASR is Mathematicoid
meager swift
Oh okay i got it
lean spade
As it turns out, you can't expand expression with rational powers in this way, for
$$(a+b)^{\frac{p}{q}}=\left[(a+b)^{p}\right]^{\frac{1}{q}}$$
vale crescentBOT
ASR is Mathematicoid
meager swift
Oh but why
lean spade
Now p is an integer, you know how to expand (a+b)^p but then you have to take q th root of the expansion, no one till today has discovered that how to expand (a+b)^1/q where q is an integer other than 1 and 0
lean spade
Oh w8, maybe I didn't knew some things about it, you can take q th root of the expansion, but those are advanced stuff🫠
lean spade
Here i have writt the formula
The condition |x|<1 ensures series convergence btw
meager swift
Convergence.?
lean spade
Convergence.?
A series is said to be convergent if it results in a finite value
meager swift
So it means the expansion does not go on to infinity?
Root 2 does not have a finite value
lean spade
So it means the expansion does not go on to infinity?
The expansion can go to infinity, the overall sum cannot
lean spade
Root 2 does not have a finite value
Root 2 doea have a finite value
meager swift
Okayy
icy hill
Root 2 does not have a finite value
No, the decimal expansion is infinite, but the number itself is finite
To say something is finite is to say it isn't infinite (or esentially, that it exists in some form of numbers)
crystal tuskBOT
@meager swift
meager swift
+close
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