Question

You must explain.
How do I solve this problem?

Answer 1

Answer:

I don't think that it is a question... this is more like a pyramid of tables from 1 to 10

if you observe

then the numbers on the left side of the triangle is the 1st multiple of each number

then move towards the right side,the numbers that are given are the consecutive multiple of the number

when reaching the very end on the right side..start reading down in left direction and that will be the other multiples of the number

For example-

take number 5

following the ways...

5 then 10 then 15 till 25 and then below it is 30 then 35 to 50

hence completing the table

Answer 2

$\large\underline{\text{Proper Question}}$

Find the sum of all numbers in the number pyramid.

$\large\underline{\text{Solution}}$

In the first row, the sum of the number is $1$.

In the second row, there are multiples of 2. The sum of the numbers is $2(1+2)$.

In the third row, there are multiples of 3. The sum of the numbers is $3(1+2+3)$.

Now, using the method, we can find the sum of every row of the pyramid.

Let's find the sum. As we know that,

$\large\red{\boxed{\red{\bold{\displaystyle\sum^{n}_{k=1}k^{3}=\left\{\dfrac{n(n+1)}{2}\right\}^{2}}}}}$

$\large\red{\boxed{\red{\bold{\displaystyle\sum^{n}_{k=1}k^{2}=\dfrac{n(n+1)(2n+1)}{6}}}}}$

$\large\red{\boxed{\red{\bold{\displaystyle\sum^{n}_{k=1}k=\dfrac{n(n+1)}{2}}}}}$

The sum of the numbers in each row is:

$\hookrightarrow1+2(1+2)+3(1+2+3)+\cdots+10(1+2+3+\cdots+10)$

We can see that each row sums to $\displaystyle k(1+2+3+\cdots+k)$. Using the summation formula to find the whole sum, we get:

$\displaystyle\sum^{10}_{k=1}k(1+2+3+\cdots+k)$

$=\displaystyle\sum^{10}_{k=1}\{k\times \dfrac{k(k+1)}{2}\}$

$=\displaystyle\sum^{10}_{k=1}\dfrac{k^{3}+k^{2}}{2}$

$=\dfrac{1}{2}\displaystyle(\sum^{10}_{k=1}k^{3}+\displaystyle\sum^{10}_{k=1}k^{2})$

$=\dfrac{1}{2}\{(\dfrac{10\times11}{2})^{2}+\dfrac{10\times11\times21}{6}\}$

$=\dfrac{1}{2}(55^{2}+5\times11\times7)$

$=\dfrac{3025+385}{2}$

$=\dfrac{3410}{2}$

$=\large{\red{1705}}\tiny{\red{\text{//}}}$

So, ② 1705 is the right answer.