JEE MATHEMATICS MCQ: MATHEMATICAL INDUCTION

1 Easy

1 of 1

Application +4 | -1

Using mathematical induction, then numbers

{a_n}{\,^\prime }s

are defined by

{a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n},\left( {n \ge 0} \right)

Then,

{a_n}

is equal to

View answer and explanation

Correct option: B

{n^3} - {n^2} + 1

Option B (

{n^3} - {n^2} + 1

) is correct because it satisfies the given recurrence relation and initial condition.

More Questions from MATHEMATICAL INDUCTION

Practice All Chapter MCQs →

\frac{1^2}{3}+\frac{1^2+2^2}{5}+\frac{1^2+2^2+3^2}{7}+\ldots

n

P\left( n \right):2 + 4 + 6 \ldots + \left( {2n} \right),\;n \in N,

\begin{array}{l} P\left( k \right) = k\left( {k + 1} \right) + 2\mathrm{implies}\\ P\left( k \right) = \left( {k + 1} \right)\left( {k + 2} \right) + 2 \end{array}

k \in N.

P\left( n \right) = n\left( {n + 1} \right) + 2

n

n

a\left( n \right) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{\left( {{2^n}} \right) - 1}}.

n(n+1)(n+2)(n+3)

n \in N

12 {~N}

8 {~m}

6 {~m}