Sigma Notation questions

Questions

• 1. Find (15Σr=5)r^3
• 2. Find (20Σr=9)6r+3
• 3. Find (10Σr=1)r(r-1)^2
• 4. Find an expression for (3nΣr=1)r^2
• 5. Factorise (nΣr=1)4r^3-12
• 6. Factorise (nΣr=1)r^2-r to the form of: kn(n+a)(n+b)
• 7. Factorise (2nΣr=n+1)r^2+7r
• 8. Use the method of differences to work out (5Σr=1)(r+1)^3 -r^3
• 9. Show that 1/2r-1 - 1/2r+1 = 2/4r^2-1 , hence find (nΣr=1)(2/4r^2-1) , hence find (nΣr=1)(1/4r^2-1) .

1, 2 , 3 and 4....

• 1. Find (15Σr=5)r^3 : Firstly we need to deal with the r=5. As we know we can (15Σr=1)r^3 - (4Σr=1)r^3. Now we can use the formula 1/4n^2(n+1)^2 and sub in both n values, so : 1/4(15)^2(15+1)^2 - 1/4(4)^2(4+1)^2 = 14300
• 2. Find (20Σr=9)6r+3 : Again, we need to deal with the r=9 first, so we can say (20Σr=1)6r+3 - (8Σr=1)6r+3. Next we simplify this further by removing the 6r and +3 from both these notations. So we can say (6(20Σr=1)r + 3(20Σr=1)1) - (6(8Σr=1)r + 3(8Σr=1)1). This is equal to (6(1/2x20(20+1)) + 3x20) - (6(1/2x8(8+1) + 3x8) which equals 1080
• 3. Find (10Σr=1)r(r-1)^ : Here we Expand r(r+1) to get r^3-2r^2+r. Now we split our sigma notation into 3 : (10Σr=1)r^3 - 2(10Σr=1)r^2 +(10Σr=1)r , which is the same as (1/4x10^2(10+1)^2) - 2(1/6x10(10+1)(20+1)) + (1/2x10(10+1)) which equals 2310.
• 4. Find an expression for (3nΣr=1)r^2 : Here we simply sub in 3n to the equation of the sum of squares equation 1/6n(n+1)(2n+1) , so 1/6 x 3n(3n+1)(6n+1) , which we can multiply 1/6 by 3n to get 1/2n or n/2. So our answer is n/2(3n+1)(6n+1).

5, 6 and 7

• 5. Factorise (nΣr=1)4r^3-12 : Rewrite as 4(nΣr=1)r^3 - 12(nΣr=1)r which is the same as 4(1/4n^2(n+1)^2) - 12(1/2n(n+1)). This can be cancelled to get n^2(n+1)^2 - 6n(n+1). Common factors are n(n+1) and whats left is [n(n+1)-6] , so n(n+1)[n(n+1)-6] , which is the same as ; n(n+1)(n^2+n-6) and factorising the quadratic gives n(n+1)(n+3)(n-2)
• 6. Factorise (nΣr=1)r^2-r to the form of: kn(n+a)(n+b) : Rewrite as (nΣr=1)r^2 - (nΣr=1)r which is the same as 1/6n(n+1)(2n+1) - 1/2n(n+1). n(n+1) is a common factor and [1/6(2n+1)-1/2] is left, so n(n+1)[1/6(2n+1)-1/2] is the same as 1/6n(n+1)[(2n+1)-3], which is the same was 1/6n(n+1)(2n-2) , which is the same as 1/3n(n+1)(n-1)
• 7. Factorise (2nΣr=n+1)r^2+7r : Rewrite as (3(2nΣr=1)r^2 + 7(2nΣr=1)r) - (3(nΣr=1)r^2 - 7(nΣr=1)r) , and use all necessary equations and simplify to get n[7n^2+15n+4]

Answers 8 and 9

• 8. -1^3 + 6^3 = 215
• 9 a). Show
• b). 2n/2n+1 (as 1 and -1/2n+1 left , so 2n+1/2n+1 - 1/2n+1 = 2n/2n+1)
• c). 1/2n+1 as it's half of 2n/2n+1