Binomial Theorem And Its Simple Applications Question 99
Question: Using mathematical induction, the numbers $ a_{n}’s $ are defined by $ a_0=1,{a_{n+1}}=3n^{2}+n+{a_{n’}} $ $ (n\ge 0). $ Then, $ a_{n} $ is equal to
Options:
A) $ n^{3}+n^{2}+1 $
B) $ n^{3}-n^{2}+1 $
C) $ n^{3}-n^{2} $
D) $ n^{3}+n^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Given, $ a_0=1,{a_{n+1}}=3n^{2}+n+a_{n} $
$ \Rightarrow a_1=3(0)+0+a_0=1 $
$ \Rightarrow a_2=3{{(1)}^{2}}+1+a_1=3+1+1=5 $
For option (b),
Let $ P(n)=n^{3}-n^{2}+1 $
$ \therefore ,P(0)=0-0+1=1=a_0 $
$ P(1)=1^{3}-1^{2}+1=1=a_1 $
and $ P(2)={{(2)}^{3}}-{{(2)}^{2}}+1=5=a_2 $
$ \therefore ,a_{n}=n^{3}-n^{2}+1 $
BETA
