JEE MATHEMATICS MCQ: MATHEMATICAL INDUCTION

1 Medium

1 of 1

Application +4 | -1

1+4+13+40+\ldots

upto

n

terms is equal to

View answer and explanation

Correct option: B

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

Option B is correct because substituting n=1 into the expression yields the correct sum, and the formula is derived by finding the general term and summing the geometric progression.

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

upto

n

terms is equal to If

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

then

\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}

is true for all

k \in N.

So, statement

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

is true for The value of sum in the

n

th bracket of For a positive integer

n

, Let

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

then The greatest positive integer, which divides

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

for all

n \in N

, is Forces forming a couple are of magnitude

12 {~N}

each and the arm of the couple is

8 {~m}

. The force of an equivalent couple whose arm is

6 {~m}

is of magnitude