
Which is greater – the sum of the first 50 whole numbers or the product of the first 100 whole numbers ?
Answer
216.3k+ views
Hint:- As we know that whole numbers begin with zero and go till infinity. And in general if all the numbers are positive then the product of those numbers will be greater than their sum but here the numbers start from zero(included).
Complete step-by-step answer:
Now as we know that whole numbers are those numbers which are perfect positive integers or we can say that all natural numbers are whole numbers and zero is also a whole number but not a natural number.
So, now let us first find the sum of the first 50 whole numbers.
As we know that the first 50 natural numbers are 0, 1, 2, 3, 4, 5, …….. , 49.
It is obvious that addition of zero to any number does not change the result .
So, we had to find the sum of whole numbers from 1 to 49 (i.e. 1, 2, 3, 4, …… , 49)
As we can see that all the numbers from 1 to 49 are positive.
Hence, the resultant sum must be a positive integer.
But now let us find the product of the first 100 whole numbers.
So, the first 100 whole numbers will be 0, 1, 2, 3, 4, 5, …….. , 98, 99.
Now as we know that if we multiply zero with any number then the resultant product will always be zero.
Like 0*x = 0, where x can be any finite number.
So, in the product of the first 100 whole numbers zero is also present.
So, the resultant product must also be zero.
Now as we know that the sum of the first 50 whole numbers is a positive integer and the product of the first 100 whole numbers is zero.
And as we know that all positive integers are greater than zero.
Hence, the sum of the first 50 whole numbers is greater than the product of the first 100 whole numbers.
Note:- Whenever we come up with this type of question then we can also find the sum of consecutive numbers using formula because as we know that zero does not affect the resultant sum. So, the sum of the first 50 whole numbers (0 to 49) will be the same as the sum of the first 49 natural numbers (1 to 49). And the formula for the sum of first n natural numbers is \[\dfrac{{n\left( {n + 1} \right)}}{2}\].
So, sum will be \[\dfrac{{49\left( {49 + 1} \right)}}{2} = 1225\]. But here we can directly say that the sum of the first 50 whole numbers will be greater because 0 multiplied with any number gives the result as zero.
Complete step-by-step answer:
Now as we know that whole numbers are those numbers which are perfect positive integers or we can say that all natural numbers are whole numbers and zero is also a whole number but not a natural number.
So, now let us first find the sum of the first 50 whole numbers.
As we know that the first 50 natural numbers are 0, 1, 2, 3, 4, 5, …….. , 49.
It is obvious that addition of zero to any number does not change the result .
So, we had to find the sum of whole numbers from 1 to 49 (i.e. 1, 2, 3, 4, …… , 49)
As we can see that all the numbers from 1 to 49 are positive.
Hence, the resultant sum must be a positive integer.
But now let us find the product of the first 100 whole numbers.
So, the first 100 whole numbers will be 0, 1, 2, 3, 4, 5, …….. , 98, 99.
Now as we know that if we multiply zero with any number then the resultant product will always be zero.
Like 0*x = 0, where x can be any finite number.
So, in the product of the first 100 whole numbers zero is also present.
So, the resultant product must also be zero.
Now as we know that the sum of the first 50 whole numbers is a positive integer and the product of the first 100 whole numbers is zero.
And as we know that all positive integers are greater than zero.
Hence, the sum of the first 50 whole numbers is greater than the product of the first 100 whole numbers.
Note:- Whenever we come up with this type of question then we can also find the sum of consecutive numbers using formula because as we know that zero does not affect the resultant sum. So, the sum of the first 50 whole numbers (0 to 49) will be the same as the sum of the first 49 natural numbers (1 to 49). And the formula for the sum of first n natural numbers is \[\dfrac{{n\left( {n + 1} \right)}}{2}\].
So, sum will be \[\dfrac{{49\left( {49 + 1} \right)}}{2} = 1225\]. But here we can directly say that the sum of the first 50 whole numbers will be greater because 0 multiplied with any number gives the result as zero.
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