Question

求数列前n项和：1+2*3+3*7+4*15+^+n*(2^n-1)

Answer 1

原式=1×2+2×4+3×8+···+n×2^n-(1+2+3+4+```+n)······①

设Sn= 1×2+2×4+3×8+···n×2^n

2Sn= 1×4+2×8+···+(n-1)×2^n+n×2^(n+1)

∴Sn=(n-1)×2^(n+1)+2······②

剩下的你会算了吧······