Question

show that twenty times the sum of first 20 even natural numbers is 21 times the sum of the first 20 odd numbers.

Answer 1

Answer:

Step-by-step explanation: first 20 even numbers are in a.p

whose 1st term (a)=2

and common difference (d)=2

∴Sum of the 20 even numbers=[20/2{4+(20-1)2}] =10(4+38)

=420

∴20 times the sum off the numbers = 20*420  =8400   =21*400                =21[20/2(2+(20-1)2}]

=21*sum of first 20 odd numbers

Answer 2

Answer:  Showed.

Step-by-step explanation:  We are given to show that twenty times the sum of first 20 even natural numbers is 21 times the sum of the first 20 odd numbers.

The sum of first 'n' even natural numbers is n(n + 1)

and the sum of first 'n' odd natural numbers is n².

So, the sum of first 20 even natural numbers is given by

$S_e=20(20+1)=20\times 21.$

And the sum of first 20 odd natural numbers is

$S_o=20^2.$

Therefore,

$20\times S_e=20\times 20\times 21=21\times 20^2=21\times S_o.$

Hence showed.