If x is positive and x^2+1/x^2=23, find x^7+1/x^7.
Answer and solution
Correct answer: 57965
Solution
Answer: 57965
1. Because x>0, x+1/x is positive. Its square is 23+2=25, so P₁=x+1/x=5.
2. Define Pn=xⁿ+x⁻ⁿ. Multiplying Pn−1 by x+1/x gives Pn+Pn−2; hence Pn=5Pn−1−Pn−2.
3. Start P₀=2, P₁=5. Compute P₂=23, P₃=110, P₄=527 and P₅=2525.
4. P₆=5×2525−527=12098. P₇=5×12098−2525=57965.
5. Independent identity check: P₃P₄=P₇+P₁, so P₇=110×527−5=57965.
Why this method works:
The identity (x+1/x)²=x²+1/x²+2 determines P₁=5 because x is positive. To derive the shortcut, expand (x³+x⁻³)(x⁴+x⁻⁴): it equals x⁷+x⁻⁷+x+x⁻¹, hence P₇=P₃P₄−P₁.
Independent check and common trap:
P₃=P₁³−3P₁=125−15=110 and P₄=P₂²−2=529−2=527. Thus 110×527−5=57965. This uses only two smaller powers and provides a separate check on the recurrence arithmetic.
Apply: P₃=5(23−1)=110; P₄=23²−2=527. Then P₇=P₃×P₄−P₁=110×527−5=57,965. This skips P₅ and P₆.
Check: P₃=P₁³−3P₁=125−15=110 and P₄=P₂²−2=529−2=527. Thus 110×527−5=57965. This uses only two smaller powers and provides a separate check on the recurrence arithmetic.
1. Because x>0, x+1/x is positive. Its square is 23+2=25, so P₁=x+1/x=5.
2. Define Pn=xⁿ+x⁻ⁿ. Multiplying Pn−1 by x+1/x gives Pn+Pn−2; hence Pn=5Pn−1−Pn−2.
3. Start P₀=2, P₁=5. Compute P₂=23, P₃=110, P₄=527 and P₅=2525.
4. P₆=5×2525−527=12098. P₇=5×12098−2525=57965.
5. Independent identity check: P₃P₄=P₇+P₁, so P₇=110×527−5=57965.
Why this method works:
The identity (x+1/x)²=x²+1/x²+2 determines P₁=5 because x is positive. To derive the shortcut, expand (x³+x⁻³)(x⁴+x⁻⁴): it equals x⁷+x⁻⁷+x+x⁻¹, hence P₇=P₃P₄−P₁.
Independent check and common trap:
P₃=P₁³−3P₁=125−15=110 and P₄=P₂²−2=529−2=527. Thus 110×527−5=57965. This uses only two smaller powers and provides a separate check on the recurrence arithmetic.
G Strategy shortcut
Spot: Algebraic Identities.Apply: P₃=5(23−1)=110; P₄=23²−2=527. Then P₇=P₃×P₄−P₁=110×527−5=57,965. This skips P₅ and P₆.
Check: P₃=P₁³−3P₁=125−15=110 and P₄=P₂²−2=529−2=527. Thus 110×527−5=57965. This uses only two smaller powers and provides a separate check on the recurrence arithmetic.
G Strategy: Spot: Algebraic Identities. Apply: P₃=5(23−1)=110; P₄=23²−2=527. Then P₇=P₃×P₄−P₁=110×527−5=57,965. This skips P₅ and P₆. Check: P₃=P₁³−3P₁=125−15=110 and P₄=P₂²−2=529−2=527. Thus 110×527−5=57965. This uses only two smaller powers and provides a separate check on the recurrence arithmetic.