A.16(4^2k-1+3^k+1)-13×3^k+1

B.4×4^2k+9×3^k

C.(4^2k-1+3^k+1)+15×4^2k-1+2×3^k+1

D.3(4^2k-1+3^k+1)-13×4^2k-1

A.16(4^2k-1+3^k+1)-13×3^k+1                    B.4×4^2k+9×3^k

C.(4^2k-1+3^k+1)+15×4^2k-1+2×3^k+1               D.3(4^2k-1+3^k+1)-13×4^2k-1

A.16（4^2k-1+3 ^k+1）-13×3^k+1

B.4×4^2k+9×3^k

C.(4^2k-1+3^k+1)+15×4^2k-1+2×3^k+1

D.3(4^2k-1+3^k+1)-13×4^2k-1

用數(shù)學(xué)歸納法證明“4^2n-1+3ⁿ⁺¹(n∈N+)能被13整除”的第二步中，當(dāng)n＝k+1時(shí)為了使用歸納假設(shè)，對(duì)4^2k+1+3^k+2變形正確的是

1. A.
   16(4^2k-1+3^k+1)-13×3^k+1
2. B.
   4×4^2k+9×3^k
3. C.
   (4^2k-1+3^k+1)+15×4^2k-1+2×3^k+1
4. D.
   3(4^2k-1+3^k+1)-13×4^2k-1

A.16(4^2k-1+3^k+1)-13×3^k+1

B.4×4^2k+9×3^k

C.(4^2k-1+3^k+1)+15×4^2k-1+2×3^k+1

D.3(4^2k-1+3^k+1)-13×4^2k-1