If ${\mathrm{u}}_{\mathrm{n}+1}=3{\mathrm{u}}_{\mathrm{n}}-2{\mathrm{u}}_{\mathrm{n}-1}$ and ${\mathrm{u}}_0=2,{\mathrm{u}}_1=3$ then ${\mathrm{u}}_{\mathrm{n}}$ s equal to

detailed solution

Correct option is B

∵un+1=3un−2un−1----istep I : Given u1=3=2+1=21+1 which is true for n =1 , Put n =1in Eq. (i), u1+1=3u1−2u1−1u2=3u1−2u0=3⋅3−2⋅2=5=22+1which is true for n =2. Therefore, the result are true for n = 1 and n = 2step II : Assume it is true for n = &, then it is also true for n =k-1.  uk=2k+1---iuk−1=2k−1+1---iiStep lll : On putting n = k in Eq. (i), we get         uk+1=3uk−2uk−1=32k+1−22k−1+1 [from Eqs. (ii) and (iii)] =3⋅2k+3−2⋅2k−1−2=3⋅2k+3−2k−2=(3−1)2k+1=2⋅2k+1=2k+1+1This shows that the result is true for n =k + 1, hence by the principle of mathematical induction the result is true for all n∈N