00:01
Hi, in this question, given that ai of x equals 1 plus summation n equals 1 to infinity x power 3n divided by 2 into 3, 5 into 6 etc up to 3n minus 1 into 3n and also given that bi of x equals x plus summation n equals 1 to infinity x power 3n plus 1 divided by 3 into 4 into 6 into 7 etc 3n into 3n plus 1.
00:47
So here an can be written as x power 3n divided by 2 into 3, 5 into 6 etc up to 3n minus 1 into 3n.
01:01
So here by using the ratio test limit n tends to infinity modulus of an plus 1 divided by an which is equal to limit n tends to infinity an plus 1 as x power 3n plus 3 divided by 2 into 3, 5 into 6 etc up to 3 into n plus 1 minus 1 into 3 into n plus 1 multiplied by so we have to write it in the reciprocal form so we can write it as x power 3n in the power we get 2 into 3, 5 into 6 etc up to 3n minus 1 into 3n.
01:49
On further simplifying which is equal to limit n tends to infinity x cube divided by 3n plus 2 into 3n plus 3.
02:03
On further simplifying which is equal to limit n tends to infinity x cube divided by 3n plus 2 into 3n minus 1 into 3n plus 2 into 3n minus 1 into 3n plus 2 into 3n minus 1 into 3n minus 1 into 3n plus 1.
02:53
So here limit n tends to infinity bn plus 1 divided by bn which is equal to limit n tends to infinity here we get x power 3 into n plus 1 plus 1 divided by 3 into 4, 6 into 7 etc up to 3 into n plus 1 into 3 into n plus 1 plus 1 multiplied by here we get 3 into 4, 6 into 7 etc up to 3n into 3n plus 1 the whole divided by x power 3n plus 1...