Question

Use induction to prove that 0–0 4j3 = n4 + 2n3 + n2 where n > 0.

Answer 1

solution 2,410 nt & 2n3a na 20 Base case: 20 insi s 4 jo = 400)3 = 4X0 =0 RHS s nt+ an3+ n2 = 0+0+020 LAS IRMS =0 . It is true for n = 0 some. e. Z 43 let it be there for the R4+ 2 R² + R² ... (1) jo Now we'll prone it for Rt be. (R +12° +Q0R+123 + (R+19 f=0 { 4 + 4CR+13 ...(2) RH 2 : 3 E 43 R RA Els : E 4j 20

substitua È 4 from eance to eqn (2) =0 RA 3 3 » 73 19 :. 341 Rad 2 K ² + R² & 4(R+1) R 4 + 2k3 t R² + 4 (R3 4 1 4 3R² + 3k) = Rt + Q R 3 + k ²4 483 4 4 + 12k ²4 12k = kat 6R3 + 13R²+12R 4 q ...(3) swvide this by (R+ 12 l.e. R21 26 +1 R 22 418 + 4 R2+ 2 k tl RA & 6k3 & 13 k² + 12 kt 4 AR4 -2 R3 - R² 46 + 12 R2 + 12 R + 4 4R3 of 8 R² of 4R 4 ke +86 + 4 - Ak? 8 R 44 Atay RH LI egn (3) can be written as 433 = (R+0 4R2+ 4R+4) J=0 (RH1 2. 2 [(k+102 + 2CR +1)],+1 =(k+1) 4 7 2 (R41)3 + (R+1) 2 . It is there for all values of n30 Hence proved

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