The sum sum_{k=1}^{20}(1+2+3+ldots+k) is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The sum is given by $\sum_{k=1}^{20} \frac{k(k+1)}{2}$.Using the formula for the sum of the first $n$ natural numbers,$1+2+\ldots+k = \frac{k(k+1)

Vedclass · Accepted Answer

$1540$

Vedclass · Answer

(C) The sum is given by $\sum_{k=1}^{20} \frac{k(k+1)}{2}$.Using the formula for the sum of the first $n$ natural numbers,$1+2+\ldots+k = \frac{k(k+1)}{2}$.We need to calculate $\frac{1}{2} \sum_{k=1}^{20} (k^2 + k)$.Using the summation formulas $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ for $n=20$:$\sum_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = 2870$.$\sum_{k=1}^{20} k = \frac{20(21)}{2} = 210$.Therefore,the total sum is $\frac{1}{2} (2870 + 210) = \frac{1}{2} (3080) = 1540$.

The sum $\sum_{k=1}^{20}(1+2+3+\ldots+k)$ is

Similar Questions

Let $a_{1}, a_{2}, a_{3}, \ldots$ be an $A.P.$ If $\sum_{r=1}^{\infty} \frac{a_{r}}{2^{r}}=4$,then $4 a_{2}$ is equal to

$\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j 1 } } = \dots$

$2 + 4 + 7 + 11 + 16 + \dots$ to $n$ terms =

The sum of the series: $(2)^2 + 2(4)^2 + 3(6)^2 + \dots$ up to $10$ terms is

$\sum_{n=1}^5 n(n^2+n+1) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module