【简答题】定义一种新运算*,满足n*k=nλ k-1 (n,k∈N * λ为非零常数). (1)对于任意给定的k,设a n =n*k(n=1,2,3,…),证明:数列{a n }是等差数列; (2)对于任意给定的n,设b k =n*k(k=1,2,3…),证明:数列{b k }是等比数列; (3)设c n =n*n(n=1,2,3,..),试求数列{c n }的前n项和S n .
【简答题】设数列{a n }:1,-2,-2,3,3,3,-4,-4,-4,-4,…,(-1) k -1 k,…,(-1) ,即当 (k∈N * )时,a n =(-1) k -1 k,记S n =a 1 +a 2 +…+a n (n∈N * ),用数学归纳法证明S i(2i +1) =-i(2i+1)(i∈N * ).