bài 1
.Ta có :
Sn=1+2a+3a2+...+nan−1Sn=1+2a+3a2+...+nan−1
→aSn=a+2a2+3a3+...+nan→aSn=a+2a2+3a3+...+nan
→aSn−Sn=−1−a−a2−a3−..−an−1+nan→aSn−Sn=−1−a−a2−a3−..−an−1+nan
→aSn−Sn=−(1+a+a2+a3+..+an−1)+nan→aSn−Sn=−(1+a+a2+a3+..+an−1)+nan
→(a−1)Sn=−an−1a−1+nan→(a−1)Sn=−an−1a−1+nan
→Sn=−an−1(a−1)2+nana−1