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).

Answer 1

證明：(1)當(dāng)n=1時(shí)，左式=1·2²-2·3²=-14，

右式=-1·2·7=-14，等式成立.

(2)假設(shè)當(dāng)n=k時(shí)等式成立，即有(1·2²-2·3²)+…+［(2k-1)(2k)²-2k(2k+1)²］=-k(k+1)(4k+3);

當(dāng)n=k+1時(shí)，有(1·2²-2·3²)+…+［(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+1)+1］·［4(k+1)+3］,

即n=k+1時(shí)等式成立.

由(1)(2)知等式對(duì)任何自然數(shù)n都成立.