问学习之路团队一道数学题目我要问的是第8题Sn=b1b2-b2b3+b3b4 -…+(-1)^n-1bnbn+1.当n是偶数时sn=-2/9n(n+3)当n是奇数时,sn=1/9(2n^2+6n+7)
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①当n为奇数时,
Sn=b1b2-b2b3+b3b4-b4b5+…+(-1)^(n-1)bnbn+1
=b1b2-b2b3+b3b4-b4b5+…+bnbn+1
=b1b2+b3(b4-b2)+b5(b6-b4)+…+bn(bn+1-bn-1)
=(1*5/3)+4(b3+b5+…+bn)
=5/3+4*[7/3+(2n+1)/3]*(n-1)/2/2
=5/3+(7+2n+1)(n-1)/3
=5/3+(8+2n)(n-1)/3
=5/3+(8n+2n^2-8-2n)/3
=(2n^2+6n-3)/3
②当n为偶数时,
b1b2-b2b3+b3b4-b4b5+…+(-1)^(n-1)bnbn+1
=b2(b1-b3)+b4(b3-b5)+…+bn(bn-1-bn+1)
=-4(b2+b4+…+bn)
=-4*[5/3+(2n+1)/3]*n/2/2
=-(5+2n+1)*n/3
=-(2n^2+6n)/3