Use the properties of summation and the Summation Formulas...

Use the properties of summation and the Summation Formulas Theorem to evaluate the sum. Use the summation capabilities of a graphing utility to verify your result.\ $sum_{i=1}^{20} (i - 1)^2$\ Find the sum by adding each term together. Use the summation capabilities of a graphing utility to verify your result.\ $sum_{i=1}^{6} (6i + 9)$\ Use the properties of summation and the Summation Formulas Theorem to evaluate the sum. Use the summation capabilities of a graphing utility to verify your result.\ $sum_{i=1}^{20} (6i - 9)$

Transcript

00:01 In the part a, the given series is i is equal to 120 in bracket i negative 1.

00:12 We need to evaluate this sum by using the property of summation.

00:18 So as we know that, the summation i is equal to 1 n f of i plus g of i is equal to summation i is equal to 1 n f of i the summation i is equal to 1 n g of i the summation i is equal to 1 n c o 5 and a submission i is equal to 1 n f of i negative g of i is equal to submission i is equal to n fopi negative summation i is equal to 1 n g of y and the submission i is equal to 1 n k times of fopi is equal to k times of summation i is equal to point n f of i and also we know that the submission of i is equal to 1 to n i square is equal to n times of n plus 1 multiply to twice of n plus 1 over 6 and the i is equal to 1 n of i is equal to n times of n plus 1 over 2 and the summation i is equal to 1 n of 1 is equal therefore now summation i is equal to 1 t of i negative 1 full square is equal to summation i is equal to 1 20 i square negative 2 i plus 1 1 1 so this is equal to summation i is equal to 1 to 20 i square take it twice of summation i is equal to 1 to 20 of i plus summation of i is equal to 1 20 of 1 further we have it is equal to 20 multiplied 21 plus 20 to the 40 plus 1 41 over 20 to the 40 plus 1.

02:33 41 over 6 negative twice of 20 multiply to 21 over 2 plus 20 plus 20 so this is equal to 2 3 jazeja 3 7 jr so it will give 2 870 negative 420 plus 20 this is equal to 2 4 4 7 0 this is ever required sum.

03:23 Now in part v, we need to find the sum, i is equal to 1, 6, 6 i plus 9.

03:34 So this is equal to summation, i is equal to 6 times of i plus summation i is equal to 1 to 6 of 9.

03:48 So this is equal to 6 multiple of 6 multiple that is 7 over 2 plus 9 multiple of 6.

04:00 This is equal to 2 3 is a 126 plus 24.

04:08 After summing we have 0 8 1...

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