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# Find the coefficient of x^(-16) in the expansion of (x^2 - 1/x)^25?

Find the coefficient of x^(-16) in the expansion of (x^2 - 1/x)^25

Please show different methods of solving this, thanks in advance

### 5 Answers

- VamanLv 711 months ago
I think that there is no answer. Any power odf(x^2)^n/x^m can give this value.

- Anonymous11 months ago
wolfram alpha

https://www.wolframalpha.com/input/?i=%28x%5E2+-+1...

x^50 - 25 x^47 + 300 x^44 - 2300 x^41 + 12650 x^38 - 53130 x^35 + 177100 x^32 - 480700 x^29 + 1081575 x^26 - 1/x^25 - 2042975 x^23 + 25/x^22 + 3268760 x^20 - 300/x^19 - 4457400 x^17 + 2300/x^16 + 5200300 x^14 - 12650/x^13 - 5200300 x^11 + 53130/x^10 + 4457400 x^8 - 177100/x^7 - 3268760 x^5 + 480700/x^4 + 2042975 x^2 - 1081575/x

+ 2300/x^16 = 2300

pari/gp calculator (x^2-1/x)^25

%1 = (x^75 - 25*x^72 + 300*x^69 - 2300*x^66 + 12650*x^63 - 53130*x^60 + 177100*x^57 - 480700*x^54 + 1081575*x^51 - 2042975*x^48 + 3268760*x^45 - 4457400*x^42 + 5200300*x^39 - 5200300*x^36 + 4457400*x^33 - 3268760*x^30 + 2042975*x^27 - 1081575*x^24 + 480700*x^21 - 177100*x^18 + 53130*x^15 - 12650*x^12 + 2300*x^9 - 300*x^6 + 25*x^3 - 1)/x^25

+ 2300*x^9 = 2300

x^9/x^25

- Steve ALv 711 months ago
The expansion terms will be (x^2)^25 = x^50, then (x^2)^24(x^-1) = x^47, etc. So each goes down by 3.

(53-(-16))/3 = 69/3 = 23

You need the 23d term, which would be the (x^2)^3(x^-1)^22 term.

25!/(22!3!) = 2300

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- rotchmLv 711 months ago
(a+b)ⁿ = ∑ nCi * a^i * b^(n-i). Your n is 25.

Letting a = x² & b = x^(-1) then a^i * b^(n-i) = (x²)^i * (x^(-1))^(25-i) = [left for you] and must equal -16.

Solve for i. What is i ?

Once found, the coefficient you seek is 25Ci.

Done!