Given a quadratic seq: $$ 0 ; 7 $$; 32 ; ... 3.1 Det the expression for the general temp Tn of the quadaticseq (4) 3.2 Which term(s) in the quadratic sequence have a value of 56 ? (3) 3.3 tence or othorwise; (alculate the unlue of $$ \sum_{n=5}^{10} T n-\sum_{n=11}^{15} $$ Tn

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**3.1 Finding the General Term**

Assume the quadratic sequence has general term  
$$
T(n)=an^2+bn+c.
$$
We are given:
- $$ T(1)= a(1)^2 + b(1) + c = a+b+c = 0, $$
- $$ T(2)= 4a+2b+c = 7, $$
- $$ T(3)= 9a+3b+c = 32. $$

Subtract the equation for $$ n=1 $$ from that for $$ n=2 $$:
$$
(4a+2b+c) - (a+b+c)= 3a+b = 7. \tag{1}
$$

Subtract the equation for $$ n=2 $$ from that for $$ n=3 $$:
$$
(9a+3b+c) - (4a+2b+c)= 5a+b = 25. \tag{2}
$$

Now, subtract equation (1) from equation (2):
$$
(5a+b) - (3a+b) = 25 - 7 \quad\Longrightarrow\quad 2a = 18,
$$
so
$$
a = 9.
$$

Substitute $$ a=9 $$ into equation (1):
$$
3(9) + b = 7 \quad\Longrightarrow\quad 27+b = 7,
$$
which gives
$$
b = -20.
$$

Finally, use $$ T(1)=0 $$ to find $$ c $$:
$$
9 - 20 + c = 0 \quad\Longrightarrow\quad c = 11.
$$

Thus, the general term is:
$$
T(n)=9n^2-20n+11.
$$

---

**3.2 Finding the Term(s) Equal to 56**

We need to solve:
$$
9n^2-20n+11=56.
$$
Subtract 56 from both sides:
$$
9n^2-20n-45=0.
$$

Using the quadratic formula, 
$$
n=\frac{-(-20)\pm\sqrt{(-20)^2-4\cdot 9\cdot(-45)}}{2\cdot 9},
$$
we find:
$$
n=\frac{20\pm\sqrt{400+1620}}{18}=\frac{20\pm\sqrt{2020}}{18}.
$$

Since 
$$
\sqrt{2020} = 2\sqrt{505},
$$
this becomes:
$$
n=\frac{20\pm2\sqrt{505}}{18} = \frac{10\pm\sqrt{505}}{9}.
$$

The positive solution is:
$$
n=\frac{10+\sqrt{505}}{9}\approx\frac{10+22.494}{9}\approx\frac{32.494}{9}\approx3.61.
$$
Since $$ n $$ must be a positive integer, there is no term in the sequence exactly equal to 56.

---

**3.3 Calculating $$ \displaystyle \sum_{n=5}^{10}T(n)-\sum_{n=11}^{15}T(n) $$**

Recall that
$$
T(n)=9n^2-20n+11.
$$
Thus,
$$
\sum_{n=a}^{b}T(n) = 9\sum_{n=a}^{b}n^2 -20\sum_{n=a}^{b}n + 11\,(b-a+1).
$$

**For $$ \displaystyle \sum_{n=5}^{10}T(n) $$:**

- Number of terms: $$10-5+1=6$$.
- Sum of $$ n $$ from 5 to 10:
  $$
  \sum_{n=5}^{10} n = \frac{10\cdot 11}{2}-\frac{4\cdot 5}{2} = 55-10=45.
  $$
- Sum of $$ n^2 $$ from 5 to 10:
  $$
  \sum_{n=5}^{10} n^2 = \left[\frac{10\cdot11\cdot21}{6}\right]-\left[\frac{4\cdot5\cdot9}{6}\right]=\frac{2310}{6}-\frac{180}{6}=385-30=355.
  $$

Therefore,
$$
\sum_{n=5}^{10}T(n)=9\cdot355-20\cdot45+11\cdot6 = 3195-900+66 = 2361.
$$

**For $$ \displaystyle \sum_{n=11}^{15}T(n) $$:**

- Number of terms: $$15-11+1=5$$.
- Sum of $$ n $$ from 11 to 15:
  $$
  \sum_{n=11}^{15} n = \frac{15\cdot16}{2} - \frac{10\cdot11}{2} = 120-55=65.
  $$
- Sum of $$ n^2 $$ from 11 to 15:
  $$
  \sum_{n=11}^{15} n^2 = \left[\frac{15\cdot16\cdot31}{6}\right]-\left[\frac{10\cdot11\cdot21}{6}\right].
  $$
  Compute each part:
  $$
  \frac{15\cdot16\cdot31}{6} = \frac{7440}{6} = 1240, \quad \frac{10\cdot11\cdot21}{6} = \frac{2310}{6} = 385.
  $$
  So,
  $$
  \sum_{n=11}^{15} n^2 = 1240-385=855.
  $$

Thus,
$$
\sum_{n=11}^{15}T(n)=9\cdot855-20\cdot65+11\cdot5 = 7695-1300+55 = 6450.
$$

**Difference:**
$$
\sum_{n=5}^{10}T(n)-\sum_{n=11}^{15}T(n)= 2361-6450=-4089.
$$

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**Final Answers:**

3.1 $$\displaystyle T(n)=9n^2-20n+11$$.  
3.2 No term of the sequence equals 56 (since the solution for $$ n $$ is nonintegral).  
3.3 $$\displaystyle \sum_{n=5}^{10}T(n)-\sum_{n=11}^{15}T(n)=-4089$$.
**3.1 General Term:** $$ T(n) = 9n^2 - 20n + 11 $$ **3.2 Terms Equal to 56:** No term in the sequence equals 56. **3.3 Difference of Sums:** $$ \sum_{n=5}^{10} T(n) - \sum_{n=11}^{15} T(n) = -4089 $$

Given a quadratic seq: \( 0 ; 7 \); 32 ; ... 3.1 Det the expression for the general temp Tn of the quadaticseq (4) 3.2 Which term(s) in the quadratic sequence have a value of 56 ? (3) 3.3 tence or othorwise; (alculate the unlue of \( \sum_{n=5}^{10} T n-\sum_{n=11}^{15} \) Tn

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