1
$\begingroup$

If the sum of $n$ terms of an A.P. is $C_n^2$.Then wht is the sum of the square of $n$ terms?I can't get the nerve of this.Hint is just needed.

$\endgroup$
0
$\begingroup$

Hint: Let the arithmetic progression be in the form $x_k=h+kd$ for $k\in\{0,\ldots,n\}$. Note that this inherently assumes the terms of the arithmetic progression you choose are consecutive.

I now make several observations: $C_0^2=\sum\limits_{k=0}^0x_k=x_0=h$, $C_1^2=\sum\limits_{k=0}^1x_k=x_0+x_1=2h+d$, which means we have $h=C_0^2$, and $d=C_1^2-2C_0^2$

Now

$$\sum\limits_{k=0}^nx_k^2=\sum\limits_{k=0}^n(h+kd)^2=\sum\limits_{k=0}^nh^2+2hkd+(kd)^2=h^2\sum\limits_{k=0}^n1+2hd\sum\limits_{k=0}^nk+d^2\sum\limits_{k=0}^nk^2$$

This gives you enough information to express $\sum\limits_{k=0}^nx_k^2$ as a function of $C_0^2$, $C_1^2$, and $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy