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8. If a, b, c and d in any binomial expansion be the 6th, 7th, 8th and 9th termsb-ac 4arespectively, then prove that |
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Answer» (x + y)^n a = 6th term = C(n,5) x^(n−5) y^5b = 7th term = C(n,6) x^(n−6) y^6c = 8th term = C(n,7) x^(n−7) y^7d = 9th term = C(n,8) x^(n−8) y^8 =================================== b^2 − ac= [C(n,6) x^(n−6) y^6]^2 − C(n,5) x^(n−5) y^5 * C(n,7) x^(n−7) y^7= [n!/(6! (n−6)!)]^2 x^(2n−12) y^12 − n!/(5!(n−5)!) * n!/(7!(n−7)!) x^(2n−12) y^12= x^(2n−12) y^12 (n!/(5!(n−7)!))^2 [1/(36(n−6)²) − 1/(42(n−5)(n−6))]= x^(2n−12) y^12 (n!/(5!(n−7)!))^2 [(n+1)/(252(n−6)²(n−5))] c^2 − bd= [C(n,7) x^(n−7) y^7]^2 − C(n,6) x^(n−6) y^6 * C(n,8) x^(n−8) y^8= [n!/(7! (n−7)!)]^2 x^(2n−14) y^14 − n!/(6!(n−6)!) * n!/(8!(n−8)!) x^(2n−14) y^14= x^(2n−14) y^14 (n!/(6!(n−8)!))^2 [1/(49(n−7)²) − 1/(56(n−6)(n−7))]= x^(2n−14) y^14 (n!/(6!(n−8)!))^2 [(n+1)/(392(n−7)²(n−6))] ------------------------------ (b^2 − ac) / (c^2 − bd) = x^(2n−12) y^12 (n!/(5!(n−7)!))^2 [(n+1)/(252(n−6)²(n−5))]————————————————————————— =x^(2n−14) y^14 (n!/(6!(n−8)!))^2 [(n+1)/(392(n−7)²(n−6))] x^2 (6!(n−8)!)^2 (392(n−7)²(n−6))———————————————— =y^2 (5!(n−7)!)^2 (252(n−6)²(n−5)) x^2 (6)^2 (98(n−7)²)————————————— =y^2 (n−7)^2 (63(n−6)(n−5)) x^2 (36) (98)————————— =y^2 (63(n−6)(n−5)) 56 x^2———————(n−6)(n−5) y^2 =================================== 4a/(3c) = 4 C(n,5) x^(n−5) y^5—————————— =3 C(n,7) x^(n−7) y^7 4 n! / (5! (n−5)!) x^2—————————— =3 n! / (7! (n−7)!) y^2 4 (7! (n−7)!) x^2———————— =3 (5! (n−5)!) y^2 4 (7*6) x^2———————— =3 (n−5)(n−6) y^2 56 x^2——————— = (b^2 − ac) / (c^2 − bd)(n−5)(n−6) y^2 |
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