Question

(2,4,6.....(2n))=2^n(1,2,3....n)​

Answer 1

Answer:

The sum of even numbers from 2 to infinity can be obtained easily, using Arithmetic Progression as well as the sum of all natural numbers. We know that the even numbers are the numbers, which are completely divisible by 2. They are 2, 4, 6, 8,10, 12,14, 16 and so on. Now, we need to find the total of these numbers. Also, find sum of odd numbers here.

Basically, the formula to find the sum of even numbers is n(n+1), where n is the natural number. We can find this formula using the formula of the sum of natural numbers, such as:

S = 1 + 2+3+4+5+6+7…+n

S= n(n+1)/2

To find the sum of consecutive even numbers, we need to multiply the above formula by 2. Hence,

Se = n(n+1)

Let us derive this formula using AP.

Sum of Even Numbers Formula Using AP

Let the sum of first n even numbers is Sn

Sn = 2+4+6+8+10+…………………..+(2n) ……. (1)

By Arithmetic Progression, we know, for any sequence, the sum of numbers is given by;

Sn=1/2×n[2a+(n-1)d] ……..(2)

Where,

n = number of digits in the series

a = First term of an A.P

d= Common difference in an A.P

Therefore, if we put the values in equation 2 with respect to equation 1, such as;

a=2 , d = 2

Let, last term, l = (2n)

So, the sum will be:

Sn = ½ n[2.2+(n-1)2]

Sn = n/2[4+2n-2]

Sn = n/2[2+2n]

Sn = n(n+1)

Sum of n even numbers = n(n+1)

Sum of First Ten Even numbers

Below is the table for the sum of the first ten even numbers.

Number of consecutive even numbers (n) Sum of even numbers (Sn = n (n+1)) Recheck

1 1(1+1)=1×2=2 2

2 2(2+1) = 2×3 = 6 2+4 = 6

3 3(3+1)=3×4 = 12 2+4+6 = 12

4 4(4+1) = 4 x 5 = 20 2+4+6+8=20

5 5(5+1) = 5 x 6 = 30 2+4+6+8+10 = 30

6 6(6+1) = 6 x 7 = 42 2+4+6+8+10+12 = 42

7 7(7+1) = 7×8 = 56 2+4+6+8+10+12+14 = 56

8 8(8+1) = 8 x 9 = 72 2+4+6+8+10+12+14+16=72

9 9(9+1) = 9 x 10 = 90 2+4+6+8+10+12+14+16+18=90

10 10(10+1) = 10 x 11 =110 2+4+6+8+10+12+14+16+18+20=110

Answer 2

Answer:

Let P(n) :2 + 4 + 6+ …+2 n = n2 + n

P(l): 2 = l2 + 1 = 2, which is true

Hence, P(l) is true.

Let us assume that P(n) is true for some natural number n = k.

∴ P(k): 2 + 4 + 6 + .,.+2k = k2 + k (i)

Now, we have to prove that P(k + 1) is true.

P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)

= k2 + k + 2(k+ 1) [Using (i)]

= k2 + k + 2k + 2

= k2 + 2k+1+k+1

= (k + 1)2 + k+ 1