Assertion $(A)$: $1+(1+2+4)+(4+6+9)+(9+12+16)+\ldots+(81+90+100)=1000$
Reason $(R)$: $\sum_{r=1}^n(r^3-(r-1)^3)=n^3$ for any natural number $n$.

  • A
    Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
  • B
    Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A)$
  • C
    $(A)$ is true but $(R)$ is false
  • D
    $(A)$ is false but $(R)$ is true

Explore More

Similar Questions

The value of $\lim _{n \rightarrow \infty} \left( \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \right)$ is:

If ${t_n} = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1, 2, 3, \dots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \dots + \frac{1}{t_{2003}} = $

Difficult
View Solution

If ${x_1}, {x_2}, {x_3}, \dots, {x_n}$ are in $A.P.$ whose common difference is $\alpha$,then the value of $\sin \alpha (\sec {x_1} \sec {x_2} + \sec {x_2} \sec {x_3} + \dots + \sec {x_{n-1}} \sec {x_n}) = $

If $\sum\limits_{n = 1}^5 {\frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} = \frac{k}{3}} $,then $k$ is equal to

If $a_n = \frac{-2}{4n^2 - 16n + 15}$,then $a_1 + a_2 + \dots + a_{25}$ is equal to:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo