JEE MATHEMATICS MCQ: MATHEMATICAL INDUCTION

1 Medium

1 of 1

Analyzing +4 | -1

\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+\ldots n

terms is equal to

View answer and explanation

Correct option: A

n-\frac{1}{3} 4^n-\frac{1}{3}

Option B (

n+\frac{{1}}{{3}} 4^{-n}-\frac{{1}}{{3}}

) is correct because substituting n=1 into the expression yields the first term of the series, 3/4.

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}