Is there a sequence?

We can rewrite the sequence as

$(\frac{4-1}{4\cdot1}+\frac{7-4}{7\cdot4}+\frac{10-7}{10\cdot7}+...+\frac{100-97}{100\cdot97})$

= $(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100})$

Everything in the middle cancels out and we're left with

$1-\frac{1}{100}$

= $\frac{99}{100}$

= $0.99$

Read more on telescoping series here

Since 3 is multiplied by the whole sequence.... The answer must be divisible by 3....only. 0.99 is divisible... Here..

S = $\frac{3}{1*4}$ + $\frac{3}{4*7}$ + $\frac{3}{7*10}$ + $\frac{3}{10*13}$ +.....+ $\frac{3}{97*100}$ then by splitting them into partial fractions , we can telescope : S = ( $\frac{1}{1}$ - $\frac{1}{4}$ + $\frac{1}{4}$ - $\frac{1}{7}$ + $\frac{1}{7}$ - ...... + $\frac{1}{97}$ - $\frac{1}{100}$ ). Thus everything cancels out except for the first and last term: S = 1 - $\frac{1}{100}$ = $\frac{99}{100}$ = 0.99