Is there an easy way?

the rule for consecutive integers is = $\frac{n(n+1)}{2}$

so, the summation is= $\frac{50 \times 51}{2}$ = $1275$

We can use the formula

$(n^2*n)/2$

which will initially equal to 1275

Answer is $\dfrac{50 × 51}{2} = \color{#69047E}{\boxed{1275}}$ .

S=

1 + 2 + 3 +... +25+

50+49 +48 +...+26

If we write this down in this order we can see that S consists of 25 pairs of numbers, each giving us the sum of 51, so S=25*51=1275

To find the sum of first natural numbers there is a formula. Let n be the last term. The formula is n(n+1)/2. By substituting the values, we find the answer.

The proof for formula:

${ s }_{ n }=1+2+3+4+......+n=\frac { n(n+1) }{ 2 } \\ { s }_{ n+1 }=1+2+3+........n+1\\ \quad \quad \quad ={ s }_{ n }\quad +\quad n+1\\ Assuming\quad { s }_{ n }\quad is\quad true\\ { s }_{ n+1 }=\frac { n(n+1) }{ 2 } +n+1\\ \quad \quad \quad =(n+1)(\frac { n }{ 2 } +1)\\ \quad \quad \quad =(n+1)(\frac { n+2 }{ 2 } )\\ \quad \quad \quad =\frac { (n+1)(n+2) }{ 2 }$ Which is exactly what we wanted for n+1.