Question

7+9+11+..+(2n+1)=an^2+bn+c a+b+c-?​

Answer 1

В левой части имеем сумму $m$ членов арифметической прогрессии, где $a_1=7$ - первый член, $d=a_2-a_1=9-7=2$ - разность прогрессии.

По формуле общего члена прогрессии найдем номер последнего члена записанной суммы:

$a_m=a_1+d(m-1);\\\\2n+1=7+2(m-1);\\\\2n-6=2m-2;\\\\2m=2n-4;\\\\m=n-2$

Т.е. в левой части записана сумма $n-2$ членов прогрессии.

По формуле $S_m=\frac{a_1+a_m}{2}\cdot m$ находим сумму в левой части равенства:

$S_{n-2}=\frac{7+2n+1}{2}\cdot(n-2)=\frac{8+2n}{2}\cdot(n-2)=(4+n)(n-2)=n^2+2n-8.$

Итого $n^2+2n-8=an^2+bn+c$, откуда $a=1$, $b=2$, $c=-8$

$a+b+c=1+2-8=-5.$

ОТВЕТ: -5.

Answer 2

The given series is an arithmetic sequence with a common difference of 2. It starts with 7 and has a pattern of adding 2 to each subsequent term. The general form of an arithmetic sequence is:

$a_n = a_1 + (n-1)d,$

where $a_n$ is the $n$th term, $a_1$ is the first term (which is 7 in this case), $n$ is the term number, and $d$ is the common difference (which is 2 in this case).

The sum of the first $n$ terms of an arithmetic sequence can be calculated using the formula:

$S_n = \frac{n}{2} \left(2a_1 + (n-1)d\right).$

In your case, $a_1 = 7$ and $d = 2$. Substituting these values into the formula:

$S_n = \frac{n}{2} \left(2 \cdot 7 + (n-1) \cdot 2\right) = n(7 + 2n).$

So, the sum of the series $7+9+11+\ldots+(2n+1)$ is $S_n = n(7 + 2n)$.

Now, comparing this to the given quadratic expression $an^2 + bn + c$, we can see that $a = 2$, $b = 7$, and $c = 0$. Therefore, $a + b + c = 2 + 7 + 0 = 9$.