Sovi.AI - AI Math Tutor

Scan to solve math questions
ENFRDEESJAZH-CNZH-TW

QUESTION IMAGE

use pascals triangle to expand each binomial. 1. \\((2 + x)^4\\) 2. \\(…

Question

use pascals triangle to expand each binomial.

  1. \\((2 + x)^4\\)
  2. \\((n + m)^5\\)
  3. \\((4a - b)^3\\)
  4. \\((x + y)^6\\)
  5. \\((3x + 2y)^7\\)
  6. \\((a - 4)^6\\)
  7. \\((3c - d)^4\\)
  8. \\((m - n)^5\\)
  9. \\((a - b)^3\\)
  10. \\((3p - 2q)^4\\)

find the coefficient of the indicated term in each expansion.

  1. \\((x - 2)^{10}\\), 5th term
  2. \\((4m + 1)^8\\), 3rd term
  3. \\((x + 3y)^{10}\\), 8th term
  4. \\((2c - d)^{12}\\), 6th term
  5. \\((a + b)^8\\), 4th term
  6. \\((2a + 3b)^{10}\\), 5th term
  7. \\((x - y)^9\\), 6th term
  8. \\((x + y)^{12}\\), 7th term
  9. \\((x + 2)^7\\), 4th term
  10. \\((a - 3)^8\\), 5th term
  11. \\((2a + 3b)^{10}\\), \\(a^6 b^4\\) term
  12. \\((2x + 3y)^9\\), \\(x^6 y^3\\) term
  13. \\(\left(x + \frac{1}{3}\

ight)^7\\), 4th term

  1. \\(\left(x - \frac{1}{2}\

ight)^{10}\\), 6th term

  1. \\((x + 4y)^7\\), \\(x^2 y^5\\) term
  2. \\((3x + 5y)^{10}\\), \\(x^6 y^4\\) term

Explanation:

Expand binomials using Pascal's triangle (Questions 1-10)

$$ LATEXBLOCK0 $$

Find the coefficient of the indicated term (Questions 11-20)

\[
\begin{aligned}
\text{General term of } (A+B)^n \text{ is } & T_{k+1} = \binom{n}{k} A^{n-k} B^k \\

  1. \quad (x-2)^{10}, \text{ 5th term } (k=4) &\implies T_5 = \binom{10}{4} x^6 (-2)^4 = 210 \cdot 16 \cdot x^6 = 3360x^6 \implies \text{Coefficient} = 3360 \\
  2. \quad (4m+1)^8, \text{ 3rd term } (k=2) &\implies T_3 = \binom{8}{2} (4m)^6 (1)^2 = 28 \cdot 4096 \cdot m^6 = 114688m^6 \implies \text{Coefficient} = 114688 \\
  3. \quad (x+3y)^{10}, \text{ 8th term } (k=7) &\implies T_8 = \binom{10}{7} x^3 (3y)^7 = 120 \cdot 2187 \cdot x^3 y^7 = 262440x^3y^7 \implies \text{Coefficient} = 262440 \\
  4. \quad (2c-d)^{12}, \text{ 6th term } (k=5) &\implies T_6 = \binom{12}{5} (2c)^7 (-d)^5 = 792 \cdot 128 \cdot (-1) \cdot c^7 d^5 = -101376c^7d^5 \implies \text{Coefficient} = -101376 \\
  5. \quad (a+b)^8, \text{ 4th term } (k=3) &\implies T_4 = \binom{8}{3} a^5 b^3 = 56a^5b^3 \implies \text{Coefficient} = 56 \\
  6. \quad (2a+3b)^{10}, \text{ 5th term } (k=4) &\implies T_5 = \binom{10}{4} (2a)^6 (3b)^4 = 210 \cdot 64 \cdot 81 \cdot a^6 b^4 = 1088640a^6b^4 \implies \text{Coefficient} = 1088640 \\
  7. \quad (x-y)^9, \text{ 6th term } (k=5) &\implies T_6 = \binom{9}{5} x^4 (-y)^5 = 126 \cdot (-1) \cdot x^4 y^5 = -126x^4y^5 \implies \text{Coefficient} = -126 \\
  8. \quad (x+y)^{12}, \text{ 7th term } (k=6) &\implies T_7 = \binom{12}{6} x^6 y^6 = 924x^6y^6 \implies \text{Coefficient} = 924 \\
  9. \quad (x+2)^7, \text{ 4th term } (k=3) &\implies T_4 = \binom{7}{3} x^4 (2)^3 = 35 \cdot 8 \cdot x^4 = 280x^4 \implies \text{Coefficient} = 280 \\
  10. \quad (a-3)^8, \text{ 5th term } (k=4) &\implies T_5…

Answer:

No.Answer
2\(n^5 + 5n^4m + 10n^3m^2 + 10n^2m^3 + 5nm^4 + m^5\)
3\(64a^3 - 48a^2b + 12ab^2 - b^3\)
4\(x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\)
5\(2187x^7 + 10206x^6y + 20412x^5y^2 + 22680x^4y^3 + 15120x^3y^4 + 6048x^2y^5 + 1344xy^6 + 128y^7\)
6\(n^6 - 24n^5 + 240n^4 - 1280n^3 + 3840n^2 - 6144n + 4096\)
7\(81c^4 - 108c^3d + 54c^2d^2 - 12cd^3 + d^4\)
8\(m^5 - 5m^4n + 10m^3n^2 - 10m^2n^3 + 5mn^4 - n^5\)
9\(a^3 - 3a^2b + 3ab^2 - b^3\)
10\(81p^4 - 216p^3q + 216p^2q^2 - 96pq^3 + 16q^4\)
11\(3360\)
12\(114688\)
13\(262440\)
14\(-101376\)
15\(56\)
16\(1088640\)
17\(-126\)
18\(924\)
19\(280\)
20\(5670\)
21\(1088640\)
22\(145152\)
23\(\frac{35}{27}\)
24\(-\frac{63}{8}\)
25\(21504\)
26\(95681250\)