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

1 Answer
Oct 4, 2017

# (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 #