The sum to $n$ terms of the series ${1^3} + {3^3} + {5^3} + \ldots $ is

${n^2}\left( {{n^2} - 1} \right)$

${n^2}\left( {{n^2} + 1} \right)$

JEE MATHEMATICS MCQ: MATHEMATICAL INDUCTION

1 Easy

1 of 1

Application +4 | -1

The sum to

n

terms of the series

{1^3} + {3^3} + {5^3} + \ldots

View answer and explanation

Correct option: B

{n^2}\left( {2{n^2} - 1} \right)

Option B (

{n^2}(2{n^2} - 1)

) is correct because it gives the sum of the series for n = 1, 2, etc.

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\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}