数列﹛an﹜满足4an-1+3,a1=0,则此数列的第5项是数列﹛an﹜满足an=4an-1+3,a1=0,则此数列的第5项是
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数列﹛an﹜满足4an-1+3,a1=0,则此数列的第5项是数列﹛an﹜满足an=4an-1+3,a1=0,则此数列的第5项是
数列﹛an﹜满足4an-1+3,a1=0,则此数列的第5项是
数列﹛an﹜满足an=4an-1+3,a1=0,则此数列的第5项是
数列﹛an﹜满足4an-1+3,a1=0,则此数列的第5项是数列﹛an﹜满足an=4an-1+3,a1=0,则此数列的第5项是
an=4an-1+3 ,n≥2
设an-λ=4(an-1 -λ)
an+1=4an-3λ
解得λ=-1
所以{an +1}是公比为4的GP
an +1=4^(n-1)
即an=4^(n-1)-1
所以a5=4^4-1=255
递推公式没写好
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