Question

用數(shù)學(xué)歸納法證明

(1×2²-2×3²)+(3×4²-4×5²)+…+［(2n-1)(2n)²-2n(2n+1)²］=-n(n+1)(4n+3)(n∈N^*).

Answer 1

證明:(1)當(dāng)n=1時(shí),左邊=1×2²-2×3²=-14,右邊=-1×2×7=-14,等式成立.?

(2)假設(shè)當(dāng)n=k時(shí)等式成立,即

(1×2²-2×3²)+(3×4²-4×5²)+…+［(2k-1)(2k)²-2k(2k+1)²］=-k(k+1)(4k+3),則當(dāng)n=k+1時(shí),

(1×2²-2×3²)+(3×4²-4×5²)+…+［(2k-1)(2k)²-2k(2k+1)²］+［(2k+1)(2k+2)²-(2k+2)(2k+3)²］

=-k(k+1)(4k+3)-2(k+1)［(4k²+12k+9)-(4k²+6k+2)］?

=-k(k+1)(4k+3)-2(k+1)(6k+7)

=-(k+1)(4k²+15k+14)?

=-(k+1)(k+2)(4k+7)?

=-(k+1)［(k+1)+1］［4(k+1)+3］.?

這說明當(dāng)n=k+1時(shí),等式也成立.?

由(1),(2)可知等式對(duì)n∈N^*都成立.

溫馨提示

用數(shù)學(xué)歸納法證明恒等式,關(guān)鍵是在證明n=k+1時(shí)命題成立.從n=k+1時(shí)的待證恒等式的一端“拼湊”出歸納假設(shè)恒等式的一端，再運(yùn)用歸納假設(shè)即可.