若數(shù)列的通項(xiàng)為分段函數(shù)或幾個(gè)特殊數(shù)列通項(xiàng)的和或差的組合等形式,則求和時(shí)可用分組轉(zhuǎn)化法,就是對(duì)原數(shù)列的通項(xiàng)進(jìn)行分解,分別對(duì)每個(gè)新的數(shù)列進(jìn)行求和后再相加減.
[典例] (2019·吉林調(diào)研)已知數(shù)列{an}是等比數(shù)列,a1=1,a4=8,{bn}是等差數(shù)列,b1=3,b4=12.
(1)求數(shù)列{an}和{bn}的通項(xiàng)公式;
(2)設(shè)cn=an+bn,求數(shù)列{cn}的前n項(xiàng)和Sn.
[解] (1)設(shè)數(shù)列{an}的公比為q,由a4=a1q3得8=1×q3,所以q=2,所以an=2n-1.
設(shè){bn}的公差為d,由b4=b1+3d得12=3+3d,所以d=3,所以bn=3n.
(2)因?yàn)閿?shù)列{an}的前n項(xiàng)和為==2n-1,數(shù)列{bn}的前n項(xiàng)和為b1n+d=3n+×3=n2+n,
所以Sn=2n-1+n2+n.
[方法技巧]
分組轉(zhuǎn)化法求和的常見(jiàn)類型
[提醒] 某些數(shù)列的求和是將數(shù)列轉(zhuǎn)化為若干個(gè)可求和的新數(shù)列的和或差,從而求得原數(shù)列的和,注意在含有字母的數(shù)列中對(duì)字母的討論.
[針對(duì)訓(xùn)練]
(2018·焦作四模)已知{an}為等差數(shù)列,且a2=3,{an}前4項(xiàng)的和為16,數(shù)列{bn}滿足b1=4,b4=88,且數(shù)列{bn-an}為等比數(shù)列.
(1)求數(shù)列{an}和{bn-an}的通項(xiàng)公式;
(2)求數(shù)列{bn}的前n項(xiàng)和Sn.
解:(1)設(shè){an}的公差為d,因?yàn)?/span>a2=3,{an}前4項(xiàng)的和為16,
所以解得
所以an=1+(n-1)×2=2n-1.
設(shè){bn-an}的公比為q,