Gauss solved his teacher's challenge of summing the integers from 1 to 100 immediately by realizing that the normal sum (1 + 2 + ... + 100) and the reverse sum (100 + 99 + ... + 1) would be equal. Adding these sums together results in 101 multiplied by the number of terms (100), giving the solution of 5050. This insight showed that mechanical solving is not as useful as understanding the underlying patterns in mathematical problems.

1. Sequences and Series CBSE, Class XI

2. A Story About Gauss When he was a child, to keep the class occupied, his teacher asked to sum all integers from 1 thru 100 Gauss solved it immediately. How?

4. Lessons from Gauss' Story Mechanical solving is not very useful Insight is key to solving mathematical problems Deutsche Mark is no longer in circulation. Euros replace Marks. Gauss ideas still havent lost their currency. (His contributions still get high marks!) While Money has currency , a Good Idea has permanence

9. Fibonacci Sequence: Ratio of Terms Sequence = 1,1,2,3,5,8,13,21,34,55 Ratio(k) = Term(k+1)/Term(k) r(1) = 1 / 1 = 1 r(2) = 2 / 1 = 2 r(3) = 3 / 2 = 1.5 r(4) = 5 / 3 = 1.67 r(5) = 8 / 5 = 1.6 r(6) = 13 / 8 = 1.625 r(7) = 21/13 = 1.615 r(8) = 34/21 = 1.619 r(9) = 55/34 = 1.617

10. Fibonacci Sequence Ratios: Converging Value Lets say ratios converge to unknown value φ So F(n+1)/F(n) = φ = F(n)/F(n-1) 1/ φ = [F(n-1)/F(n)] F(n+1) = F(n)+F(n-1) F(n+1)/F(n) = 1 + [F(n-1)/F(n)] φ = 1 + 1/ φ or φ * φ = φ + 1 or ) φ = (1 + sqrt(5))/2 φ is irrational and often called Golden Ratio Ancient Greek fascination. Parthenon length/ht ~= φ φ = 1.6180339887...

12. Arithmetic Progressions A sequence where, other than first term every term = previous term + common difference Notation: a = first term L = last term d = common difference S(n) = sum to n terms (Note: Slightly different notation from NCERT text)

13. General Term of Arithmetic Progression 1 st Term = a = a + ( 1 -1)*d 2 nd Term = a + d = a + ( 2 -1)*d 3 rd Term = a + 2d = a + ( 3 -1)*d kth Term = a + ( k -1)*d If the series has n terms, then Last term is nth term and so last term = L = a + ( n -1)*d

14. Example: First + Last Term = 80 for an AP with large no. of terms What is (a) Sum of 2nd term + last but 1 term (b) Sum of 3rd term + last but 2 term 2nd Term = First Term + common diff. Last but 1 Term = Last Term – common diff. 2 nd + Last but 1 Term = First Term + Last Term 3rd Term = 2nd Term + common diff. Last but 2 Term = Last but 1 Term – common diff. 3rd + Last but 2 Term = 2 nd Term + Last but 1 Term = First Term + Last Term So answer must be 80 for both questions.

16. Example: A.N. Kolmogorov at Age 5 Worked out sum of first n odd numbers for school journal first odd number = 1 = first term 2nd odd number = 3 = 1 + (2-1)*2 nth odd number = 1 + (n-1)*2 = last term 1,3,5,7,...till 1+(n-1)*2 is an AP with common difference 2 Summing this series yields an interesting answer Sum = [ first term + last term ]*(number of terms)/2 [1 + {1+(n-1)*2}]*n/2 = [ 1+1 +2n -2 ]*n/2 = n*n 1+3 = 2*2; 1+3+5=3*3; 1+3+5+7=16=4*4; 1+3+5+7+9=25=5*5

24. Geometric Progressions (GP) AP: any term = previous term + common difference (d) GP: any term = previous term * common ratio (r) Above defines AP and GP for all terms except the first AP: Kth term = (k-1)th term + d GP: Kth term = (k-1)th term * r Examples of GP: 1,2,4,8,16,32,64 Common ratio (r) = 2 1,1/2,1/4,1/8,1/16/,1/32 Common ratio (r) = 1/2