Question

已知{a_n}为等差数列，且a_n≠0，公差d≠0．
（1）试证：；；；
（2）根据（1）中的几个等式，试归纳出更一般的结论，并用数学归纳法证明．

Answer 1

解：（1）证明：由{a_n}为等差数列可得a_n-a_n-1=d（n≥2），则1/a_{1}-1/a_{2}=a_{2}-a_{1}/a_{1a_{2}}=d/a_{1a_{2}}得证；{C/^{0_{2}}}{a_{1}}-{C/^{1_{2}}}{a_{2}}+{C/^{2_{2}}}{a_{3}}=1/a_{1}-2/a_{2}+1/a_{3}=1/a_{1}-1/a_{2}+1/a_{3}-1/a_{2}=a_{2}-a_{1}/a_{1a_{2}}+a_{2}-a_{3}/a_{2a_{3}}=d•a_{3}-a_{1}/a_{1a_{2}a_{3}}=2d²/a_{1a_{2}a_{3} }得证；{C/^{0_{3}}}{a_{1}}-{C/^{1_{3}}}{a_{2}}+{C/^{2_{3}}}{a_{3}}-{C/^{3_{3}}}{a_{4}}=1/a_{1}-3/a_{2}+3/a_{3}-1/a_{4}=（1/a_{1}-1/a_{4}）-（3/a_{2}-3/a_{3}）=3d/a_{1a_{4}}-3d/a_{2a_{3}}=3d•a_{2}a_{3}-a_{1}a_{4} /a_{1a_{2}a_{3}a_{4}}=3d•2 d　/a_{1a_{2}a_{3}a_{4}}=6d³/a_{1a_{2}a_{3}a_{4}}得证．（2）结论：{C/^{0_{n-1}}}{a_{1}}-{C/^{1_{n-1}}}{a_{2}}+{C/^{2_{n-1}}}{a_{3}}-…+(-1)^{n+1/C{^{n-1}_{n-1}}}{a_{n}}=(n-1)!d^{n-1}/a_{1a_{2}…a_{n}}，证：①当n=2，3，4时，等式成立，②假设当n=k时，{C/^{0_{k-1}}}{a_{1}}-{C/^{1_{k-1}}}{a_{2}}+{C/^{2_{k-1}}}{a_{3}}-+(-1)^{k+1/C{^{k-1}_{k-1}}}{a_{k}}=(k-1)!d^{k-1}/a_{1a_{2}a_{k}}成立，那么当n=k+1时，因为C_kⁱ-1=C_k-1ⁱ-1+C_k-1^k-2，所以{C/^{0_{k}}}{a_{1}}-{C/^{1_{k}}}{a_{2}}+{C/^{2_{k}}}{a_{3}}-+(-1)^{k+2/C{^{k}_{k}}}{a_{k+1}}={C/^{0_{k-1}}}{a_{1}}-{C/^{1_{k-1}}+{C}{⁰_{k-1}}}{a_{2}}+{C/^{2_{k-1}}+{C}{¹_{k-1}}}{a_{3}}-+(-1)^{k+1}({C/^{k-1_{k-1}}+{C}{^{k-2}_{k-1}})}{a_{k}}+(-1)^{k+2/C{^{k-1}_{k-1}}}{a_{k+1}}=({C/^{0_{k-1}}}{a_{1}}-{C/^{1_{k-1}}}{a_{2}}+{C/^{2_{k-1}}}{a_{3}}-+(-1)^{k+1/C{^{k-1}_{k-1}}}{a_{k}})-({C/^{0_{k-1}}}{a_{2}}-{C/^{1_{k-1}}}{a_{3}}+{C/^{2_{k-1}}}{a_{4}}-+(-1)^{k+1/C{^{k-1}_{k-1}}}{a_{k+1}})=(k-1)!d^{k-1}/a_{1a_{2}a_{k}}-(k-1)!d^{k-1}/a_{2a_{3}a_{k}}=(k-1)!d^{k-1}/a_{1a_{2}a_{k}}(a_{k+1}-a_{1})=k!d^{k}/a_{1a_{2}a_{k}a_{k+1}}，所以，当n=k+1时，结论也成立．综合①②知，{C/^{0_{n-1}}}{a_{1}}-{C/^{1_{n-1}}}{a_{2}}+{C/^{2_{n-1}}}{a_{3}}-+(-1)^{n+1/C{^{n-1}_{n-1}}}{a_{n}}=(n-1)!d^{n-1}/a_{1a_{2}a_{n}}对n≥2都成立．