MCQ
If ${S_n} = \frac{{n(n + 1)\left( {n + 2} \right)}}{6}$ then $\sum\limits_{n = 1}^\infty  {\frac{1}{{{t_n}}}}  = $
  • A
    $1$
  • B
    $6$
  • $2$
  • D
    $\frac {1}{6}$

Answer

Correct option: C.
$2$
c
$\mathrm{t}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$

$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{6}-\frac{(\mathrm{n}-1) \mathrm{n}(\mathrm{n}+1)}{6}$

$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)}{6}[\mathrm{n}+2-\mathrm{n}+1]=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$

$\sum\limits_{n = 1}^\infty  {\frac{2}{{n\left( {n + 1} \right)}}}  = 2\left[ {\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3}.......} \right] = 2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$. Let $\hat{\mathrm{c}}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ is :
The area bounded by the parabola $y^2 = 4x$ and the line $2x - 3y + 4 = 0$, in square unit, is
If direction cosines of two lines are proportional to $(2, 3, -6)$ and $(3, -4, 5)$, then the acute angle between them is
The circumference of the circle $x^2 + y^2 - 2x + 8y - q = 0$ is bisected by the circle $x^2 + y^2 + 4x + 12y + p = 0$ , then $p + q$ is equal to
$\int_{}^{} {[f(x)\,g''(x) - f''(x)\,g(x)]\,dx} $=
Box $I$ contains $30$ cards numbered $1$ to $30$ and Box $II$ contains $20$ cards numbered $31$ to $50 .$ A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box $I$ is
The minimum value of ${{\log x} \over x}$ in the interval $[2,\,\infty )$ is
The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right|$ $= 11$ is
The probability of hitting a target by three marks men is $\frac{1}{2} , \frac{1}{3}$ and $\frac{1}{4}$ respectively. If the probability that exactly two of them will hit the target is $\lambda$ and that at least two of them hit the target is $\mu$ then $\lambda + \mu$ is equal to :-
The solution of $\frac{{dy}}{{dx}} = {\left( {\frac{y}{x}} \right)^{1/3}}$ is