Question

(a^5+b^5) formula I want explination also​

Answer 1

Assuming you understand the meaning of exponentiation, and the distributive property, we can derive it using some induction. Let’s start by teasing apart one of the terms

(a+b)5

(a+b)(a+b)4

a(a+b)4+b(a+b)4

So, whatever (a+b)4 is, we just multiply it by a , then, on the other side of the paper, multiply it by b , then we’ll add those two partial products.

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So, what’s (a+b)4 ?

(a+b)(a+b)3

a(a+b)3+b(a+b)3

Look familiar?

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Clearly, this will keep happening. Let’s drill down to the bottom and work our way up.

(a+b)2

(a+b)(a+b)

a(a+b)+b(a+b)

a2+ab+ab+b2

a2+2ab+b2

Now we’ll use this result as we back up a step:

(a+b)3

a(a+b)2+b(a+b)2

a(a2+2ab+b2)+b(a2+2ab+b2)

(a3+2a2b+ab2)+(a2b+2ab2+b3)

a3+3a2b+3ab2+b3

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Are you noticing the symmetry in the coefficients? I’ll throw in the implicit 1’s in the next step so you’ll see them.

(a+b)4

1a(1a+1b)3+1b(1a+1b)3

1a(1a3+3a2b+3ab2+1b3)+1b(1a3+3a2b+3ab2+1b3)

(1a4+3a3b+3a2b2+1ab3)+(1a3b+3a2b2+3ab3+1b4)

1a4+4a3b+6a2b2+4ab3+1b4

So, one way to perform this process mechanically is to take the current string of coefficients, copy it, shift that copy over one slot, then add them up. The “shifting” occurs because the exponents of the a -terms increase by one in the left-hand partial sum, and the exponents of the b -terms increase by one in the right-hand partial sum.

To wrap up dramatically, we’ll apply this abstraction to perform the process mechanically.

(a+b)5

(a+b)(a+b)4

a(a+b)4+b(a+b)4

a(1a4+4a3b+6a2b2+4ab3+1b4)+b(1a4+4a3b+6a2b2+4ab3+1b4)

abstraction+1

[1,4,6,4,1,0]+[0,1,4,6,4,1]

[1,5,10,10,5,1]

abstraction-1

1a5+5a4b+10a3b2+10a2b3+5ab4+1b5