Tell us what’s 2+3?
The answer is of course 5.
Now, tell us what’s (2+3)^2?
We can do it in two ways:
1) Either we first calculate 2+3=5
and then 5^2 which is 25.
2) OR we can just apply the formula
of (a+b)^2 which is a^2 + 2×a×b + b^2
Now you say that the first one was too simple and everyone will prefer to do it and of course it’s a right decision.
But now tell us what’s 99^2?
Now also we have two options, the one is by simple calculation and the other one is writing it as
(100-1)^2 and then applying the formula of (a+b)^2.
In this example, the second way is much better than the first.
Because here, our priority is accuracy and not time, and the second method will be easy to do so.
Now, if this method of solving problems is important, than we must know how the formula of (a+b)^2 came so that we may discover the formualas for higher powers also.
Let’s assume ourselves as ancient mathematicians or may be the ones who literally discover the formula of (a+b)^2.
Let’s try to observe the pattern first.
(2+3)^2 is 25.
Now 2^2 is 4 and 3^2 is 9.
And 4+9 is 13.It doesn’t matches the answer.
Now, 13+12 is 25.
So, we have to calculate 12 by using these two numbers.We can do it as
2×3=6 and then 2×6 which is 12.
So, we can write it as: Now, we may repeat the process for more examples and may get the formula of (a+b)^2.
Now, after getting the formula, our goal is to prove it. The proof of this formula is more than simple!! We can prove it as:
Now, after proving the above formula, we may want to find the formulas for other powers.
And by following the above steps we can easily find the formula for some other powers such as the formula for (a+b)^3 as
Now, it’s quite complicated to find the formula for every power and then remember it!!!
So, we have to talk about the general trend in these formula so we can easily find the formula for any power without performing complex calculations.
Let’s write down the formulas up to the power 4:
Now, if we observe the basic things in each formula, we may find that in each formula, the power of a starts from the the general power and keep decreasing till the last term where the power is 0.
Similarly, the power of b starts from 0 and keep increasing upto the highest one.
Now, let’s compare the coefficients as:
Look what we have!! It’s nothing but Pascal’s triangle. This is a triangle discovered by Pascal in which the top most element is 1 and then by taking the sum of the two elements, we can find the next term as shown below:
1+1=2, write it below
2+1=3, write it below
And so on!!
Now, we can easily find the next row of this triangle and can find the formula for (a+b)^5 by using the above observations.
Ok, so we have found a way to find the formula for any power.
But what if we want to find the formula for (a+b)^9 or (a+b)^19 ?
So we have to extend the triangle upto 9 or 19.
Hey!! this is really a lengthy and boring method!!
So, now our purpose is to find the desired row of Pascal’s triangle.
Let’s look at the pattern again!!
If we see the marked portion,
We may write this as
Now, if we have to find the nth term of this series, we may notice that:
So, we can easily find the nth term by using the combination formula.
Now let’s look at the other marked portion:
Now, we have 1,4,10..so on. Again, if we want to find the nth term, we can do so as: Now, we are almost done!!!! Let’s write the Pascal’s triangle in terms of combination notation: Now, we can easily find any row of Pascal’s triangle as: If power is n, then the row is:
And hence we can easily find the formula for any higher power!!
Did you know this is nothing but Binomial theorem
Now, we may define the Binomial theorem as:
For any positive integer n,
We have: Now, if talk about the general term, for example if we want to find out the 4th term of the binomial expansion (a+b)^5, then we can easily found it as
So, here 3 is used commonly in the above formula and 3 is 4-1.But for finding the general term, we don’t write it as r-1 (for the rth term) because it will get more complicated. So, instead we write
The general term as
We have not denoted it by rth term but we have denoted it by( r+1) th term to make the expression simple.
So, if we want 5th term then r is 4 and we can easily apply it to the above formula.
What if we want the middle term?
Ok!! Let’s talk about it,
Suppose we have to find the middle term of 1,2,3,4,5,6,7.
Then, the middle term is 4.
Note that we have 7 terms which is an odd number and the middle term is 4th term and we can write 4 as (7+1)/2.So, for any odd number of terms, we have the middle term as (n+1)/2.
What if we have even number of terms?
Suppose, we have 1,2,3,4,5,6
Then, 3,4 are the two middle terms.
So, we have two middle terms which are (n)/2 and (n)/2 + 1
These are the two consecutive terms.
Now, in the binomial expansion (a+b)^n,
If n is even, then we have total n+1 terms which is odd.
In every binomial expansion
(a+b)^n, we have n+1 terms.
So now, the middle term is
Which is (n/2 + 1)th term.
Now, if n is odd, then n+1 is even and we have two terms which are
2) (n+1)/2 +1
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