Find the sum $\sum\limits_{k = 1}^{100} {(3 + k)} $.
A.$300$
B.$5050$
C.$5300$
D.$5350$
E.$5400$

Hint: We will expand $\sum\limits_{k = 1}^{100} {(3 + k)} $ by putting the value of k. Thereafter, we will separate the series of k and 3, then we will use arithmetic progression to solve the series. This approach will give the answer.


Complete step by step solution:
Given:$\sum\limits_{k = 1}^{100} {(3 + k)} $
Here, we will put the value of $k = 1,2,3,.....100$
\[\sum\limits_{k = 1}^{100} {(3 + k) = (3 + 1) + (3 + 2) + (3 + 3)..... + (3 + 100)} \]
Now, we will separate the value
$\sum\limits_{k = 1}^{100} { = (3 + 3 + .... + 3) + (1,2,3 + .... + 100)} $
Here $3$ is repeated up to $100$ times, so
$\sum\limits_{k = 1}^{100} { = 3 \times 100 + (1,2,3 + .... + 100)} $
Let $A = \{ 1,2,3 + .... + 100)$
This series is an arithmetic progression, where
$a = 1$
$d = {a_2} - {a_1} = 2 - 1 = 1$
$n = 100$
${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
${S_{100}} = \dfrac{{100}}{2}[2 \times 1 + (100 - 1) \times 1]$
${S_n} = 50[2 + 99 \times 1]$
${S_n} = 50[2 + 99]$
${S_n} = 50 \times 101$
${S_n} = 5050$
So, $\sum\limits_{k = 1}^{100} {(3 + k) = 300 + 5050} $
$\sum\limits_{k = 1}^{100} {(3 + k) = 5350} $
Therefore, $\sum\limits_{k = 1}^{100} {(3 + k)} $ is $5350$
Hence, the correct option is C.

Note: Students must note that the sum of natural series is given as, $1 + 2 + 3 + .... + 100 = \dfrac{{n(n + 1)}}{2}$. If the given range is of whole numbers ,then find the sum of the series in accordance to the range given