Question

How do you expand `(w^2+1/w)^9`?

Answer 1

`(w^2+1/w)^9 = w^18 + 9 w^15 + 36 w^12 + 84 w^9 + 126 w^6 + 126 w^3 +`
`" " 84 + 36/w^3 + 9/w^6 + 1/w^9`

Explanation:

Recall the Binomial Theorem

`(a+b)^n = sum_(r=0)^n \ ( (n), (r) ) \ a^(n-r) \ b^r`

Where:

`( (n), (r) ) =""^nC_r = (n!)/((n-r)!r!`

is the combinatorial, which are also the numbers from the `nth` row of Pascal's Triangle

Thus we have:

`(w^2+1/w)^9 = sum_(r=0)^9 \ ( (9), (r) ) \ (w^2)^(9-r) \ (1/w)^r`
`" " = sum_(r=0)^9 \ ( (9), (r) ) \ (w^2)^(9-r) \ (1/w)^r`

`" " = ( (9), (0) ) (w^2)^9 \ (1/w)^0 + ( (9), (1) ) (w^2)^8 \ (1/w)^1 +`
`" " ( (9), (2) ) (w^2)^7 \ (1/w)^2 + ( (9), (3) ) (w^2)^6 \ (1/w)^3 +`
`" " ( (9), (4) ) (w^2)^5 \ (1/w)^4 + ( (9), (5) ) (w^2)^4 \ (1/w)^5 +`
`" " ( (9), (6) ) (w^2)^3 \ (1/w)^6 + ( (9), (7) ) (w^2)^2 \ (1/w)^7 +`
`" " ( (9), (8) ) (w^2)^1 \ (1/w)^8 + ( (9), (9) ) (w^2)^0 \ (1/w)^9`

`" " = (1) (w^18) \ (1) + (9) (w^16) \ (1/w) +`
`" " (36) (w^14) \ (1/w^2) + (84) (w^12) \ (1/w^3) +`
`" " (126) (w^10) \ (1/w^4) + (126) (w^8) \ (1/w^5) +`
`" " (84) (w^6) \ (1/w^6) + (36) (w^4) \ (1/w^7) +`
`" " (9) (w^2) \ (1/w^8) + (1) (1) \ (1/w^9)`

`" " = w^18 + 9 w^15 + 36 w^12 + 84 w^9 + 126 w^6 + 126 w^3 +`
`" " 84 + 36/w^3 + 9/w^6 + 1/w^9`