Sum of First n Even Natural Numbers

Example 1 :

Find the sum of

2 + 4 + 6 + ........ to 50 terms

Solution :

Using 2 + 4 + 6 + ........ to n terms = n(n + 1), 

2 + 4 + 6 + ........ + 50 terms = 50(50 + 1) 

= 50(51)

= 2550

Example 2 : 

Find the value of

2 + 4 + 6 + ........ + 90

Solution :

Find the value of n. 

n = l/2

= 90/2

= 45

2 + 4 + 6 + ........ + 90 :

= 2 + 4 + 6 + ........ to 45 terms

Using 2 + 4 + 6 + ........ to n terms = n(n + 1), 

= 45(45 + 1)

= 45(46)

= 2070

Example 3 : 

Find the value of

22 + 24 + 26 + ........ + 100

Solution :

22 + 24 + 26 + ........ + 100 :

= (2 + 4 + 6 + ........ + 100) - (2 + 4 + 6 + ........ + 20)

= (2 + 4 + 6 + .... to 50 terms) - (2 + 4 + 6 + .... to 10 terms)

Using 2 + 4 + 6 + ........ to n terms = n(n + 1), 

= 50(50 + 1) - 10(10 + 1)

= 50(51) - 10(11)

= 2550 - 110

= 2440

Example 4 : 

Using the sum of first n even natural numbers, find the sum of

6 + 12 + 18 + ........ to 50 terms

Solution :

6 + 12 + 18 + ........ to 50 terms :

= 3(2 + 4 + 6 + ........ to 50 terms)

Using 2 + 4 + 6 + ........ to n terms = n(n + 1), 

= 3[50(50 + 1)]

= 3[50(51)]

= 7650

Example 5 : 

Using the sum of first n even natural numbers, find the value of

6 + 12 + 18 + ........ + 360

Solution :

6 + 12 + 18 + ........ + 360 : 

= 3(2 + 4 + 6 + ........ + 120)

= 3(2 + 4 + 6 + ........ to 60 terms)

Using 2 + 4 + 6 + ........ to n terms = n(n + 1), 

= 3[60(60 + 1)]

= 3[60(61)]

= 10980

Example 6 : 

If 2 + 4 + 6 + ........ + k = 72, then find k. 

Solution :

Find the value of n. 

n = l/2

= k/2

2 + 4 + 6 + ........ to k/2 terms  = 72

Using 2 + 4 + 6 + ........ to n terms = n(n + 1),

(k/2)(k/2 + 1) = 72

(k/2)[(k + 2)/2] = 72

k(k + 2)/4 = 72

Multiply each side by 4.

k(k + 2) = 288

k² + 2k = 288

k² + 2k - 288 = 0

Factor and solve.

k² + 18k -16k - 288 = 0

k(k + 18) - 16(k + 18) = 0

(k + 18)(k - 16) = 0

But k ≠ -18, because k is an even natural number. 

Hence k = 16.

Example 7 :

If 2 + 4 + 6 + ........ to k terms = 240, then find k. 

Solution :

2 + 4 + 6 + ........ to k terms = 240

Using 2 + 4 + 6 + ........ to n terms = n(n + 1),

k(k + 1) = 240

k² + k = 240

k² + k - 240 = 0

Factor and solve. 

k² + 16k - 15k - 240 = 0

k(k + 16) - 15(k + 16) = 0

(k + 16)(k - 15) = 0

But k ≠ -16, because number of terms can not be negative value. 

Hence k = 15.

Example 8 :

Find the average of first 25 even natural numbers. 

Solution :

Using 2 + 4 + 6 + ........ to n terms = n(n + 1) to find the sum of first 25 odd natural numbers. 

2 + 4 + 6 + ........ to 25 terms = 25(25 + 1)

= 25(26)

= 650

Average of first 25 even natural numbers : 

= (Sum of first 25 even natural numbers)/25

= 650/25

= 26

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