题目内容
用数学归纳法证明42n+1+3n+2能被13整除,其中n∈N*.
见解析
证明:(1)当n=1时,42×1+1+31+2=91能被13整除.
(2)假设当n=k时,42k+1+3k+2能被13整除,
则当n=k+1时,
42(k+1)+1+3k+3=42k+1·42+3k+2·3-42k+1·3+42k+1·3
=42k+1·13+3·(42k+1+3k+2)
∵42k+1·13能被13整除,42k+1+3k+2能被13整除,
(2)假设当n=k时,42k+1+3k+2能被13整除,
则当n=k+1时,
42(k+1)+1+3k+3=42k+1·42+3k+2·3-42k+1·3+42k+1·3
=42k+1·13+3·(42k+1+3k+2)
∵42k+1·13能被13整除,42k+1+3k+2能被13整除,
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