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Further Pure 1

Further Pure 1. Summation of finite Series. Sigma notation. In the last lesson we met the following rules. 1) 1 + 2 + 3 + …… + n = (n/2)(n+1) 2) 1 2 + 2 2 + 3 2 + …… + n 2 = (n/6)(n+1)(2n+1) 3) 1 3 + 2 3 + 3 3 + …… + n 3 = (n 2 /4) (n+1) 2

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Further Pure 1

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  1. Further Pure 1 Summation of finite Series

  2. Sigma notation • In the last lesson we met the following rules. 1) 1 + 2 + 3 + …… + n= (n/2)(n+1) 2) 12 + 22 + 32 + …… + n2 = (n/6)(n+1)(2n+1) 3) 13 + 23 + 33 + …… + n3 = (n2/4)(n+1)2 • We can write long summations like the ones above using sigma notation.

  3. Sigma notation • The r acts as a counter starting at 1 (or whatever is stated under the sigma sign) and running till you get to n (on top of the sigma sign). • Each r value generates a term and then you simply add up all the terms. • The terms in the example above come from r = 1 2×1+1 = 3 r = 2 2×2+1 = 5 r = 3 2×3+1 = 7 r = 4 2×4+1 = 9 • The 4 on top of the sigma sign tells us to stop when r = 4.

  4. Questions • Here are some questions for you to try and find the values of.

  5. Sigma notation • We can now remember the identities that we met last lesson and have mentioned already adding the sigma notation.

  6. Using Nth terms • Use the nth term to find the following summation. • The summation only works if you sum from 1 to n. • How would you calculate the next example. • Here the sum goes from r = 4, to r = 8. • This means you do not want the terms for r = 1, 2 & 3. • So the answer will be the sum to 8 minus the sum to 3.

  7. Rules of summing series • Here are 2 rules that you need to be familiar with. • There is a numerical example followed by a general rule • k and a represent random constants.

  8. Example • These results can be used to find the sum to n of lots of different series. • First break the summation up. • Next use the general formula. • Here (n/4)(n+1) is a factor • Next just multiply out and collect up like terms. • Finally the expression will factorise.

  9. Question • Try this question

  10. Question • Here the sum starts at r = 6. • This is not as complicated as it may seem. • All you need to do is take of the first 5 terms. • So the sum from 6 to 30 is the sum to 30 minus the sum to 5.

  11. Questions • Here are some questions for you to find the nth terms of. • The solutions are on the next two slides

  12. Solutions

  13. Solutions

  14. Summation of a finite Series • When Carl Friedrich Gauss was a boy in elementary school his teacher asked his class to add up the first 100 numbers. • S100 = 1 + 2 + 3 + …………… + 100 • Gauss had a flash of mathematical genius and realised that the sum had 50 pairs of 101 • Therefore S100 = 50 × 101 = 5 050 • From this we can come up with the formula for the sum of the first n numbers. • Sn = (n/2)(n+1) • We have met this result a few times already.

  15. Method of differences • We can prove the same result using a different method. • The method of differences.

  16. Example 1 • Use the method of differences to find the sum to 30 of the following example. • Solution to part ii is on the next 2 slides. • You covered adding fractions in C2 and should be able to get the answer.

  17. Example 1 • We can use the identity to re-arrange the question. • Now write the summation out long hand. Starting with r = 1. • Then r = 2,3 etc. • Write out the last 2 or 3 terms. • Having written out the full summation you can spot that parts of the sum cancel. • The bits that are left do not cancel and we can sort out the sum.

  18. Example 2 • In this next example we will find the sum to n. • Solution to part ii is on the next 2 slides. • You covered adding fractions in C2 and should be able to get the answer.

  19. Example 2 • We can use the identity to re-arrange the question. • Now write the summation out long hand. Starting with r = 1. • Then r = 2,3 etc. • Write out the last 3 terms. • Having written out the full summation you can spot that parts of the sum cancel. • The bits that are left do not cancel and we can sort out the algebra.

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