Question

I need help with all 3 questions please thank you

Answer 1

Answer:

see below

Step-by-step explanation:

2,4,6,8,....

The sequence is defined by

an = a1 +d(n-1)

a1 is the first term which is 2

d  is the common difference

d = 4-2 =2

an = 2 + 2(n-1)

We want the 100 term

a100 = 2 + 2(100-1)

         =2 +2(99)

         =2 +198

         = 200

Next we need to find the sum

Sn = n/2 ( 2a1 + d(n-1))

n =100

Substituting n=100, d=2 and s1 = 2

S100 = 100/2 (2*2 + 2(100-1))

         =50 (4+ 2*99)

          =50 (4+198)

          =50 (202)

          =10100

Sum of 101 odd number

n= 101 a1 =1 and d =2

S101 = 101/2 ( 2*1 + 2(101-1))

         50.5 (2+2(100))

           50.5 (202)

           10201

The sum of the first 100 even numbers is smaller than the sum of the first 101 odd numbers

Answer 2

Answer:

Step-by-step explanation:

Problem One

Find the 100th term of the even natural numbers starting at 2

Formula

a_n = a + (n - 1)*d

Givens

• a_1 = 2             Given
• n = 100             Given
• d = 2                Given

Solution

a_100 = 2 + (100 - 1)*2

a_100 = 2 + 99*2

a_100 = 2 + 198

a_100 = 200

Problem Two

Formula

Sum = (a_1 + a_100)*n/2

Givens

a_1= 2

a_100 = 200

n = 100

Solution

Sum = (2 + 200)*100/2

Sum = (202)*50

Sum = 10100

Problem 3

Sum of the first 100 odd numbers

Givens

a_1 = 1

n = 101

Formula

a_101 = a_1 + (101 -  1)*2

Solution

a_100 = 1 + 100*2

a_100 = 1 + 200

a_100 = 201

Sum = (a_1 + a_100)*n/2

Sum = (1 + 201)*101/2

Sum = (202)*101/2

Sum = 101 * 101

Sum = 10201

Conclusion

The sum of the first 101 odd numbers is greater than the sum of the first 100 even numbers