Question

Use Principle of Mathematical Induction to show that for all n e N, an = 212.521 11 + 32 .221 11 is divisible by 19.

Answer 1

Let P(n) be the given statement that is P(n): an = 2+2.52n + 4 ght? 2n+1 is divisible by 19 when n=r PU: Q1 = 25'+3?2? = 103 +63 = (10+5) (10?-66+62) = 16. (100-60+36) = 16*76 = 16 x 1984 = 64819 , which is divisible by 19 so po is true Now we assume that P(m) is true that is, am is divisible by 19, where mis some positive integer, greater than 1 that since 19lam =) am = 19.4 for some KEN = 2m +2 2 2m +'+ 3 m+2 2 2m +1 = 190k Now for n=m+1 2(m+ 1)ot I me142 2 melofa 3 2 mellot I 2 amel - 7 5

mf2 2.2 52.5.2m +1 3.3992 2² 22m ft = 50.2me? 52m +1 + 12.3m+? 2 2m + 1 mt2 = 12 (2m+? 52m++ 37*222*7') + 38 mai f mf2 amfil 3 - 2 28 . 2 m+2 2m +1 .5 = 12. am 438.2me? 5.2m Al = 12.19.K + 19,2m 73.52m.pl = 19 (126+ 2043520+) - 19:42 me3 amfi where Kor 12kf 2 5 7 P(mt) is frine. whenever pom) is true Thay P(1) is true and p (mel) is true wheneves prom) is true. Hence by the principle of matheonatical indretion pon) is trime. for NEIN