Lesson Remarkable identities for Binomial Coefficients

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Remarkable identities for Binomial Coefficients


Problem 1

Prove this identity for  Binomial Coefficients

C%5Bn%5D%5E0 + C%5Bn%5D%5E1 + C%5Bn%5D%5E2 + . . . + C%5Bn%5D%5Ek + . . . + C%5Bn%5D%5En = 2%5En

for any positive integer number  n.

Solution
The proof is very straightforward.  Simply substitute  x = 1  into the  Binomial Expansion

%281+%2B+x%29%5En =   =   +sum+%28+C%5Bn%5D%5Ei%2Ax%5Ei+%2C+i=0%2C+n+%29+

(see the lesson  Binomial Theorem  under the current topic in this site).  You will get the required identity.

This is how it looks for the low order binomial coefficients:

n = 1,  C%5B1%5D%5E0 = 1,  C%5B1%5D%5E1 = 1,  C%5B1%5D%5E0 + C%5B1%5D%5E1 = 1 + 1 = 2;

n = 2,  C%5B2%5D%5E0 = 1,  C%5B2%5D%5E1 = 2,  C%5B2%5D%5E2 = 1,  C%5B2%5D%5E0 + C%5B2%5D%5E1 + C%5B2%5D%5E2 = 1 + 2 + 1 = 4;

n = 3,  C%5B3%5D%5E0 = 1,  C%5B3%5D%5E1 = 3,  C%5B3%5D%5E2 = 3,  C%5B3%5D%5E3 = 1,  C%5B3%5D%5E0 + C%5B3%5D%5E1 + C%5B3%5D%5E2 + C%5B3%5D%5E3 = 1 + 3 + 3 + 1 = 8.


Problem 2

Prove this identity for  Binomial Coefficients

C%5Bn%5D%5E0 - C%5Bn%5D%5E1 + C%5Bn%5D%5E2 + . . . + %28-1%29%5Ek%2AC%5Bn%5D%5Ek + . . . + %28-1%29%5En%2AC%5Bn%5D%5En = 0

(alternate sum) for any positive integer number  n.

Solution
Again, the proof is very straightforward.  Simply substitute  x = -1  into the  Binomial Expansion

%281+%2B+x%29%5En =   =   +sum+%28+C%5Bn%5D%5Ei%2Ax%5Ei+%2C+i=0%2C+n+%29+

(see the lesson  Binomial Theorem  under the current topic in this site).  You will get the required identity.

This is how it looks for the low order binomial coefficients:

n = 1,  C%5B1%5D%5E0 = 1,  C%5B1%5D%5E1 = 1,  C%5B1%5D%5E0 - C%5B1%5D%5E1 = 1 - 1 = 0;

n = 2,  C%5B2%5D%5E0 = 1,  C%5B2%5D%5E1 = 2,  C%5B2%5D%5E2 = 1,  C%5B2%5D%5E0 - C%5B2%5D%5E1 + C%5B2%5D%5E2 = 1 - 2 + 1 = 0;

n = 3,  C%5B3%5D%5E0 = 1,  C%5B3%5D%5E1 = 3,  C%5B3%5D%5E2 = 3,  C%5B3%5D%5E3 = 1,  C%5B3%5D%5E0 - C%5B3%5D%5E1 + C%5B3%5D%5E2 - C%5B3%5D%5E3 = 1 - 3 + 3 - 1 = 0.


Problem 3

Prove this identity for  Binomial Coefficients

C%5Bn%5D%5E0 + 2%2AC%5Bn%5D%5E1 + 2%5E2%2AC%5Bn%5D%5E2 + . . . + 2%5Ek%2AC%5Bn%5D%5Ek + . . . + 2%5En%2AC%5Bn%5D%5En = 3%5En

for any positive integer number  n.

Solution
Similar to the Problem 1 and Problem 2 above, the proof is very straightforward.  Simply substitute  x = 2  into the  Binomial Expansion

%281+%2B+x%29%5En =   =   +sum+%28+C%5Bn%5D%5Ei%2Ax%5Ei+%2C+i=0%2C+n+%29+

(see the lesson  Binomial Theorem  under the current topic in this site).  You will get the required identity.



By doing in a similar way,  you can prove many other identities for Binomial Coefficients,  like these

C%5Bn%5D%5E0 + 3%2AC%5Bn%5D%5E1 + 3%5E2%2AC%5Bn%5D%5E2 + . . . + 3%5Ek%2AC%5Bn%5D%5Ek + . . . + 3%5En%2AC%5Bn%5D%5En = 4%5En,

C%5Bn%5D%5E0 + %281%2F2%29%2AC%5Bn%5D%5E1 + %281%2F2%5E2%29%2AC%5Bn%5D%5E2 + . . . + %281%2F2%5Ek%29%2AC%5Bn%5D%5Ek + . . . + %281%2F2%5En%29%2AC%5Bn%5D%5En = %283%2F2%29%5En,

C%5Bn%5D%5E0 - %281%2F2%29%2AC%5Bn%5D%5E1 + %281%2F2%5E2%29%2AC%5Bn%5D%5E2 + . . . + %28%28%28-1%29%5Ek%29%2F2%5Ek%29%2AC%5Bn%5D%5Ek + . . . + %28%28%28-1%29%5En%29%2F2%5En%29%2AC%5Bn%5D%5En = 1%2F2%5En.


For the close subject on the  Pascal's triangle  see the lesson  The Pascal's triangle  under the current topic in this site.


My lessons on Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion in this site are

    - Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion
    - Remarkable identities for Binomial Coefficients                                                                          (this lesson)
    - The Pascal's triangle
    - Solved problems on binomial coefficients
    - Solving equations that include binomial coefficients and numbers of permutations
    - Math circle level problem on binomial coefficients
    - OVERVIEW of lessons on Binomial Expansion, Binomial coefficients and the Pascal's triangle


Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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