(x-2)^2( X-5)^12 求第x^7項的係數,常項數

(x - 2)^2 (x - 5)^12= (x^2 - 4x + 4) (x - 5)^12考慮 (x - 5)^12 展開式之通項為 T(r+1) = (12Cr) x^(12-r) (-5)^r令 r = 7 , T(8) = (12C7) x^(12-7) (-5)^7 = -61875000 x^5
令 r = 6 , T(7) = (12C6) x^(12-6) (-5)^6 = 14437500 x^6
令 r = 5 , T(6) = (12C5) x^(12-5) (-5)^5 = -2475000 x^7故 x^7 項的係數為 -61875000(1) + 14437500(-4) + -2475000(4)
= - 129525000常數項 = 4 * (- 5)^12 = 976562500