Question

Find the sum of the first 50 even natural numbers.

Answer 1

Hint: This is a question to find the sum of the first 50 even natural numbers.
That is \[{{S}_{n}}=2+4+6+--+50\]
This is an arithmetic series and we have formula for this as, \[{{S}_{n}}=\dfrac{n}{2}\left[ a+(n-1)d \right]\]
To apply this formula, we should know the values of the first number ‘a’, common difference ‘d’ and the number of terms ‘n’. We will get these values in the given series itself. Then substitute the values in the formula of Sn. After simplification of this we will get the required sum.

Complete step-by-step answer:
Write the given series, \[{{S}_{n}}=2+4+6+--+50\]
By seeing the series we can write the values of
\[a=2,n=50,d={{a}_{2}}-{{a}_{1}}\]
                                       \[=4-2=2\]
We are using the formula to find the sum of the series is
\[{{S}_{n}}=\dfrac{n}{2}\left[ a+(n-1)d \right]\]
Now substitute the values of a, n and d to the above formula,
\[{{S}_{n}}=\dfrac{50}{2}\left[ 2(2)+(50-1)2 \right]\]
     \[\begin{align}
  & =\dfrac{50\times 2}{2}\left[ 2+49 \right] \\
 & =50(51) \\
 & =2550 \\
\end{align}\]
Hence the sum of the first 50 even natural numbers is 2550.

Note: in this problem we have to find the sum of arithmetic sequences. It is an arithmetic sequence because the difference between any two consecutive terms is the same. This difference is also called a common difference. Also know the values of the first number(a) and number of terms(n) in that series. Now we can apply the formula to find the sum of series.