Physics, asked by Anonymous, 1 year ago

↪20 points

Expand using binomial theorem:
(a + b)^{6}

Answers

Answered by Anonymous
3

(

a

+

b

)

6

=

a

6

+

6

a

5

b

+

15

a

4

b

2

+

20

a

3

b

3

+

15

a

2

b

4

+

6

a

b

5

+

b

6

Why does this work?

If we write out the value as a product of binomials we have:

(

a

+

b

)

6

=

(

a

+

b

)

(

a

+

b

)

(

a

+

b

)

(

a

+

b

)

(

a

+

b

)

(

a

+

b

)

If you pick one term from each binomial and multiply them together, then you have made

6

choices of left or right. If you choose all left, then you end up with

a

6

- which you can do just one way. If you choose all right then you end up with

b

6

, which again you can only do one way.

Otherwise, you are making a mixture of left and right choices, analogous to picking your way down from the top of Pascal's triangle to the bottom, via a sequence of left and right branches. The power of

a

resulting is the number of left branches you choose and the power of

b

the number of right branches.

Each number in Pascal's triangle is the sum of the two above, each of which counts the number of ways to reach that point by a sequence of left and right choices. So all of the numbers in Pascal's triangle count the number of ways to reach them by left/right choices starting at the top.


Anonymous: yes
Answered by Swarup1998
7

Binomial theorem :

For any positive integer n,

\mathrm{(a+x)^{n}}

\mathrm{=a^{n}+^{n}C_{1}\:a^{n-1}x+^{n}C_{2}\:a^{n-2}x^{2}+...+x^{n}}

Solution :

Now, \mathrm{(a+b)^{6}}

\mathrm{=a^{6}+^{6}C_{1}a^{6-1}b+^{6}C_{2}a^{6-2}b^{2}}\\\mathrm{+^{6}C_{3}a^{6-3}b^{3}+^{6}C_{4}a^{6-4}b^{4}}\\\mathrm{+^{6}C_{5}a^{6-5}b^{5}+b^{6}}

\mathrm{=a^{6}+6a^{5}b+15a^{4}b^{2}+20a^{3}b^{2}}\\\mathrm{+15a^{2}b^{4}+6ab^{5}+b^{6}}

[ You can also check the given Pascal's triangle from the given attachment for clarification. ]

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