Question

find the sum of all odd numbers up to 30

Answer 1

The question can be solved using factorial, exponents and roots. As the sum of an odd number of odd numbers can not be even.

Clearly, if there are 2n odd numbers to be added, i.e, an even number of odd numbers, then their sum will always be even. The sum of even number of odd numbers is always even.

But, as odds and evens are opposite, that way, the sum of odd number of odd numbers can't be even.

So, as I said earlier, we can solve the question using exponents, factorials and roots, I show you 10 examples. 3 for each category and 1 special.

Solution number one.
First of all, I will begin with factorials.

Factorials come up even you multiply all the numbers before a number (excluding zero, including the number)

For example, I will use the number that we come up in my solutions. 3! (three factorial) =1*2*3 = 2*3 =6.

Let's look forward for our solutions.
1. 3! + 15 + 9
2. 3! + 13 + 11
3. 3! + (13-1) + (13-1)

Here, 3 factorial is actually not an even number, so that is some kind of cheating, right? But it's the fault of the question, not ours.

The first answer, as you can see, ends up to 6+15+9.
6+15+9 = 21 + 9 = 30.

The second answer is showing that 6+13+11=19 + 11 = 30.

So, till here, we were able to get 30. Next answer can be said as the special of this category, as it shows the rule of VBODMAS and factorials together.

So, as standard, we'll solve brackets first.

3!+12+12
Multiplication.
6+12+12
Yeah. Factorials contain Multiplication.
Addition.
18+12
= 30.

Here we move to our second category, exponents.

As earlier, firstly I will explain exponents.

Exponents are written in the form of
x
n

Here, x and n are integers, which can be equal to any integer, including zero. Here,
x
n
means (x*x*x.......) n times.

Let's again take the example of what we'll work with.

33
3
3
= 3*3*3,=9*3,=27.

Now, when we know that what are exponents, know one more thing, that these solutions may be some sort of complex.

4. 33+(1−1)+3
3
3
+
(
1
−
1
)
+
3

5. 33+(1+1)+1
3
3
+
(
1
+
1
)
+
1

6. 33+(3+3)−3
3
3
+
(
3
+
3
)
−
3


There are more possible solutions, but as I promised for 10, I'd like to give you 10.

Let's have a look at the fourth solution, which indicates 27+0+3, = 27+3 = 30

The fifth solution is more like the fourth one.
27+2+1
=29+1
=30

As earlier, the last solution is some kind of special, using exponents and VBODMAS

27+6-3
=33-3
=30
Or
27+6-3
=27+3
=30.

Coming to the last, roots.

What are roots? This explains which number has to be multiplied with itself a given number of times to get another given number.

For example, what we are going to use.

Third root of 27 means that which number had to be multiplied by itself 3 times to get 27. As 33=27
3
3
=
27
, the same way, third root of 27=3.

Let's see the answers.

7. Third root - 27 + 15 + (15-3)
8. Third root - 27 + 13 + (15-1)
9. Third root - 27 + 11 + (15+1)

Yes, there are more from the same pattern, but, I promised you.

Let's have a look at the answers.

Seventh solution states as below.

3 + 15 +12
= 18 + 12
= 30.

Till now, success is following us.
A glance on the eighth solution.

3 + 13 + 14
= 3 + 27
= 30.

Nearly the final solution, we haven't failed a single time.

3 + 11 + 16
= 14 + 16
= 30.

We hereby conclude our answers.

This would have been unexpected.

3^3 + 3

Yes, I know you were expecting a big equation, so, yeah, surprise!

Thanks.
Hope you're happy.

Answer 2

Answer:

225

Step-by-step explanation:

odd no are:

1,3,5,7.........29

This nis an AP

a=1,d=2,n=?

29=1+(n-1)*2

28=(n-1)*2

n-1=28/2=14

n=15

Sum=15/2(2+14*2)

=15/2*(30)

=15*15=225