binomial theorem by mathematical induction?

Answer 1

we know that coefficient of (a^k)(b^(n-k)) in (a+b)^n is
for k = 0 to n
nCk (this is combination symbol kinldy excuse my form)

now when n= 1 (a+b)^n = a+b

Let it be true for n we need to prove for n+1

(a+b)^(n+1)=(a+b)^n *(a+b)
coffecient of a^kb^(n+1-k)
on the LHS
= (n+1)Ck
on the rhs
if we take a from 2nd term it is coefficient of a^(k-1)b(n+1-k) from 1st term = nCk -1

if we take b from 2nd term it is coefficient of a^(k)b(n-k) from 1st term = nCk
so it is
nCk-1 + nCk whcih is (n+1)Ck

same as left side
Hence QED

Answer 2

The idea of induction is to prove something for the case of n = 0 or another integral starting point. Assume validity for n. This is called the induction hypothesis. Then prove it works for n + 1. As soon as you have the proof for n + 1, you have proven it for all integer values because if it works when n = 1, it must work for 2. When n = 2, it must work for 3. And, so on. Just plug 1 in for n and run the calculations from there.

Answer 3

Notation: C(n,k) = n!/[k!(n-k)!] is the binomial coefficient.

Showing that (a+b)^n = sum k=0 to n C(n,k)a^k b^(n-k)

Base case: n=0. (a+b)^0 = 1, and
sum k=0 to n C(n,k)a^k b^(n-k) = C(0,0)a^0 b^(0-0) = 1.

Induction step: assume the formula holds for n:
(a+b)^n = sum k=0 to n C(n,k)a^k b^(n-k)
and show it holds for n+1.
(a+b)^(n+1) = [sum k=0 to n C(n,k)a^k b^(n-k)] (a+b)
= [sum k=0 to n C(n,k)a^(k+1) b^(n-k)] +
[sum k=0 to n C(n,k)a^k b^(n+1-k)]
= [sum k=1 to n+1 C(n,k-1)a^k b^(n+1-k)] +
[sum k=0 to n C(n,k)a^k b^(n+1-k)]
(in first sum, I shifted the limits on k by +1, and replaced k by k-1 throughout, so the sum doesn't change)
= C(n,n)a^(n+1)b^0 +
[sum k=1 to n [C(n,k-1) + C(n,k)]a^k b^(n+1-k)] +
C(n,0)a^0 b^(n+1)
= C(n+1,n+1)a^(n+1)b^0 + (because C(k,k)=1 for all k)
[sum k=1 to n n![1/[(k-1)!(n+1-k)!] + 1/[k!(n-k)!]] a^k b^(n+1-k)] +
C(n,0)a^0 b^(n+1)
= C(n+1,n+1)a^(n+1)b^0 +
[sum k=1 to n n![(k+n-k+1)/[k!(n+1-k)!]] a^k b^(n+1-k)] +
C(n,0)a^0 b^(n+1)
= C(n+1,n+1)a^(n+1)b^0 +
[sum k=1 to n n![(n+1)/[k!(n+1-k)!]] a^k b^(n+1-k)] +
C(n,0)a^0 b^(n+1)
= sum k=0 to n+1 C(n+1,k) a^k b^(n+1-k)

and by induction, the proof is complete.

Answer 4

properly in fact any exponential function will at last overtake a polynomial this could be obtrusive data: for n=3 a hundred twenty five>39 assume n>3 and this has been shown for m<=n then 5^n=5^(n-a million)*5 yet by hypothesis 5^(n-a million)>13(n-a million)=13n-13 so for this reason 5^n>5(13n-13)=65n-sixty 5 now if we are in a position to coach that sixty 5(n-a million)>13n for n>3 we would have accomplished the induction assume 65n-sixty 5<=13n => 65n-13n-sixty 5<=0 =>52n<=sixty 5 for n>3 this is obviously fake

Answer 5

Because it's easier for me to visualize,
Pascal's Triangle:
1.........1 1
2........1 2 1
3.......1 3 3 1
4......1 4 6 4 1
5...1 5 10 10 5 1
6..1 6 15 20 15 6 1
1...........1C0 1C1
2........2C0 2C1 2C2
3.....3C0 3C1 3C2 3C3
4..4C0 4C1 4C2 4C3 4C4
Therefore
(a + b)^n =
a^n*b^0 + nC1(a^(n-1))*b + ...+nC(n-1))ab^(n-1) + b^n

It being true for the 1st case, the kth case and the (k+1)th case it is true for all n. (twice, once for the number triangle, and once for combinations)

Answer 6

assume until N u have proved
write out the N+1 anbd notice it also satisfied your assumption

Answer 7

(a+b)^1=a+b
(a+b)^2=a^2+2ab+b^2
=>2C0a^2+2C1a^2-1b^1+2C2a^2-2b^2
(a+b)^3=3C0a^3+3C1a^3-1b^1+3C2a^3-2b^2+3C3a^3-3b^3
lt this be true for some n=k
then (a+b)^k=kC1a^k+kC2a^k-1b^1+kC2a^k-2b^2+............................+kCra^k-rb^r+.........+kCkb^k
if it holds for k+1 the theorem is proved
(a+b)^k+1=(a+b)^k*(a+b)
(a+b)[a^k+kC1a^k-1b+kC2a^k-2b^2+..............+b^k]
kC0a^k+1+kC1a^kb+kC2a^k-1b^2+.......+kCra^k+1-rb^r+..kCkab^k
+kC0a^kb+kC1a^k-1b^2+....+kC(r-1)a^k+1b^r+....+kCkb^k+1
=(k+1)C0a^k+1+(k+1)C1a^kb+(k+1)C2a^k-1b^2+.......+(k+1)Cra^k+1-rb^r+.....+(k+1)C(k+1)b^k+1
so it holds for anyk=n and hence the theoren
(a+b)^n=a^n+nC1a^n-1b+nC2a^n-2b^2+........+nCra^n-rb^r+....+b^n