If $1^2+2^2+3^2 + ... + 10^2=385$, then what is the value of $2^2+4^2+6^2 + ... + 20^2$?
Options are $770$, $1155$, $1540$, $7890$.
I haven't tried it yet and I'm writing such extra things to meet the requirements to post the question...Plz solve the above question only
On
$1^2+2^2+3^2+...+10^2=385$
$\Rightarrow 4\times(1^2+2^2+3^2+...+10^2)=1540$
$\Rightarrow 1^2\times 4+2^2\times 4+3^2 \times 4+...+10^2 \times 4=1540$
$\Rightarrow 1^2\times 2^2+2^2 \times 2^2+3^2\times 2^2+...+10^2\times 2^2=1540$
$\Rightarrow (1\times2)^2+(2\times2)^2+(3\times2)^2+(4\times2)^2+...+(10 \times 2)^2=1540$
$\Rightarrow 2^2+4^2+6^2+8^2+...+20^2=1540$.
You can just reverse the steps above to find your answer.
You are given the equation, $$1^2 + 2^2 + 3^2 +\cdots + 10^2 = 385.$$ From this equation, you want to find the value of $x$ in the following equation: $$2^2 + 4^2 + 6^2 +\cdots + 20^2 = x.$$ I know the problem you have been given does not include a variable $x$, but we will assign the unknown expression to such a variable in order to work it out algebraically. Now let's line up both the equations: $$\begin{align} 1^2 + 2^2 + 3^2 +\cdots + 10^2 &= 385\tag1 \\ 2^2 + 4^2 + 3^2 +\cdots + 20^2 &= x.\tag2\end{align}$$ Notice that every term being squared in equation $(2)$ is twice as large as every term being squared in equation $(1)$. To make that properly clear, let's re-write equation $(2)$ as follows: $$\begin{align} 1^2 + 2^2 + 3^2 +\cdots + 10^2 &= 385\tag1 \\ (2\times 1)^2 + (2\times 2)^2 + (2\times 3)^2 +\cdots + (2\times 10)^2 &= x.\tag2\end{align}$$ Now if we divide every squared term in equation $(2)$ by $2^2$, this will definitely equal equation $(1)$, which we observe being equal to $385$. We divide by $2^2$ because each term being squared is twice as large as each term being squared in equation $(1)$, and since we are squaring them, we must divide by $2^2$. But since we are adding up all the squared terms together, then by dividing each term by $2$, we are dividing the entire sum by $2$ as well. Now we let this entire sum be equal to $x$, so now we can write that $$\begin{align} \frac{x}{2^2} &= 385 \\ \\ \frac{x}{4} &= 385.\tag{since $2^2 = 4$}\end{align}$$ Remember that whatever you do to the left hand side of an equation, you must always do the same to the right hand side as well, in order for both sides to be equal. It is a bit like balanacing a seesaw, such that if you add weight on one side, you need to add the same amount of weight on the other side for the seesaw to be levelled.
So how does this analogy help us?
Well, we want to use this analogy to solve for $x$, and we can by multiplying $4$ to both sides. Since multiplication is the opposite of division, and we are dividing $x$ by $4$, then to just get $x$ alone, we multiply by $4$. But remember: what we do the left hand side, we must also do to the right hand side. So, we multiply the right hand side by $4$ as well, which is equal to $385$. We should now have something that looks like this: $$\begin{align} \frac{x}{4}\times 4 &= 385\times 4 \\ x &= 385\times 4\end{align}$$ When something is alone on one side, this is known as the subject of the equation, and is what we want to solve for. In this case, we have assigned our subject as $x$, and we see that $x = 4\times 385$. And since $385\times 4 = 1540$, we finally conclude that $$\boxed{ \ x = 1540 \ }$$ and there is your answer!