text: 1. ## Algebra
  
  Hello,
  can any "smart" person who is very good at math solve this question? it is (d) in the attachment....
  good luck
  thanks
  
  2. $S_n$ is defined by $S_n= \frac{1}{6}n(n+1)(n+2)$ so $S_{n-1}= \frac{1}{6}(n-1)((n-1)+ 1)((n-1)+ 2)= \frac{1}{6}(n-1)(n)(n+1)$
  
  You could multiply both of those out and subtract but it is simpler to notice that each has factors of 1/6, n, and n+1 andfactor those out:
  $S_{n}- S_k{n-1}= \frac{1}{6}(n)(n+1)(n+2- (n-1))= \frac{1}{6}n(n+1)(3)$
  
  3. ## sequence oroblem
  
  Originally Posted by igcsestudent
  Hello,
  can any "smart" person who is very good at math solve this question? it is (d) in the attachment....
  good luck
  thanks
  sequence 1,4,10,20,35 Find nextterm
  Differences 3,6,10, 15, next difference is 21 6 comes after 3,4,5
  35+21 is 56
  
  bjh
  
  4. cool!
explanation: The document contains multiple posts related to math questions and their answers. The first post introduces a math question from an attachment, but the question itself is not provided in the text. The second post provides a detailed solution to a math problem involving the sequence \( S_n \). The third post introduces another math question about finding the next term in a sequence and provides the answer. The fourth post is a brief acknowledgment.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 1. ## Algebra
  
  Hello,
  can any "smart" person who is very good at math solve this question? it is (d) in the attachment....
  good luck
  thanks
  
  2. $S_n$ is defined by $S_n= \frac{1}{6}n(n+1)(n+2)$ so $S_{n-1}= \frac{1}{6}(n-1)((n-1)+ 1)((n-1)+ 2)= \frac{1}{6}(n-1)(n)(n+1)$
  
  You could multiply both of those out and subtract but it is simpler to notice that each has factors of 1/6, n, and n+1 andfactor those out:
  $S_{n}- S_k{n-1}= \frac{1}{6}(n)(n+1)(n+2- (n-1))= \frac{1}{6}n(n+1)(3)$
  
  3. ## sequence oroblem
  
  Originally Posted by igcsestudent
  Hello,
  can any "smart" person who is very good at math solve this question? it is (d) in the attachment....
  good luck
  thanks
  sequence 1,4,10,20,35 Find nextterm
  Differences 3,6,10, 15, next difference is 21 6 comes after 3,4,5
  35+21 is 56
  
  bjh
  
  4. cool!
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains multiple posts related to math questions and their answers. The first post introduces a math question from an attachment, but the question itself is not provided in the text. The second post provides a detailed solution to a math problem involving the sequence \( S_n \). The third post introduces another math question about finding the next term in a sequence and provides the answer. The fourth post is a brief acknowledgment.
  
  Label: multiple-question-and-answer