EXPAND AND EVALUATE SIGMA NOTATION

Expand and evaluate :

Problem 1 :

Solution :

k = 1 to k = 3

= (4 × 1) + (4 × 2) + (4 × 3)

= 4 + 8 + 12

= 24

Problem 2 :

Solution :

k = 1 to k = 6

= (1 + 1)  + (2 + 1 )  + (3  + 1)  + (4 + 1)  + (5 + 1)  + (6 + 1)

= 2  +  3  + 4  + 5  +  6 + 7

= 27

Problem 3 :

Solution :

k = 1 to k = 4

= (3(1) - 5)  + (3(2) - 5 )  + (3(3)  - 5)  + (3(4) - 5)

= (3 - 5) + (6 - 5) + (9 - 5) + (12 - 5)

=  -2  +  1  + 4  + 7 

= 10

Problem 4 :

Solution :

k = 1 to k = 5

= (11 - 2(1))  + (11 - 2(2))  + (11  - 2(3))  + (11 - 2(4)) + (11 - 2(5))

= (11 - 2) + (11 - 4) + (11 - 6) + (11 - 8) + (11 - 10)

=  9  +  7  + 5  + 3 + 1 

= 25

Problem 5 :

Solution :

k = 1 to k = 7

= 1(1 + 1)  + 2(2 + 1)  + 3(3  + 1)  + 4(4 + 1) + 5(5 + 1) + 6(6 + 1) + 7(7 + 1)

= 1(2) + 2(3) + 3(4) + 4(5) + 5(6) + 6(7) + 7(8)

=  2  +  6  + 12  + 20 + 30 + 42 + 56 

= 168

Problem 6 :

Solution :

k = 1 to k = 5

= 10 × 2^{1 - 1} + 10 × 2^{2 - 1} + 10 × 2^{3 - 1} + 10 × 2^{4 - 1} + 10 × 2^{5 - 1}

= 10 × 2⁰ + 10 × 2¹ + 10 × 2² + 10 × 2³ + 10 × 2⁴

= 10 × 1 + 10 × 2 + 10 × 4 + 10 × 8 + 10 × 16

= 10 [1 + 2 + 4 + 8 + 16]

= 10 [1 + 2 + 2²+ 2³ + 2⁴]

1 + 2 + 2²+ 2³ + 2⁴ it is geometric series.

= 10(31)

= 310

Problem 7 :

Solution :

k = 1 to k = 5

= 1(1 + 1) (1 + 2) +  2(2 + 1) (2 + 2) + 3(3 + 1) (3 + 2) + 4(4 + 1) (4 + 2) + 5(5 + 1) (5 + 2)

= 1(2) (3) + 2(3) (4) + 3(4) (5) + 4(5) (6) + 5(6) (7)

= 6 + 24 + 60 + 120 + 210

= 420

Problem 8 :

Solution :

= 100 × (1 .2)^{(k - 3)}