Question

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✅ Question in attachment
✅Polynomials
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✴Challenge to mods.....
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Answer 1

QUESTION 1 :

Find the middle term of (3x/7 - 2y)¹⁰.

ANSWER 1:

• - ¹⁰C₅ (6/7)⁵ x⁵ y⁵

EXPLANATION:

The expansion for the general term  (a + b)ⁿ is,

$\boxed{\bold{\large{\gray{T_{r+1}= \ ^nC_r\ a^{n-r}\ b^r}}}}$

10 in the power, also it is even

$\boxed{\bold{\large{\gray{Middle\ term = \dfrac{n + 2}{2}}}}}$

$\sf Middle\ term = \dfrac{10 + 2}{2}$

$\sf Middle\ term = \dfrac{12}{2}$

Middle term = 6

∴ 6th term is the middle term

r + 1 = 6

r = 5

n = 10

a = 3x/7

b = - 2y

$\sf T_6= \ ^{10}C_5\ \left(\dfrac{3x}{7}\right)^{10-5}\ (-2y)^5$

$\sf T_6= -\ ^{10}C_5\ \left(\dfrac{3x}{7}\right)^{5}\ 2^5y^5$

$\sf T_6= -\ ^{10}C_5\ \left(\dfrac{3}{7}\right)^{5}\ 2^5\ x^{5} \ y^5$

$\sf T_6= -\ ^{10}C_5\ \left(\dfrac{6}{7}\right)^{5}\ x^{5} \ y^5$

$\boxed{\large{\sf {Middle \ term = -\ ^{10}C_5\ \left(\dfrac{6}{7}\right)^{5}\ x^{5} \ y^5}}}$

QUESTION 2:

Find the middle term of (7p + 2q)⁸.

ANSWER 2:

• ⁸C₄ (14)⁴ (pq)⁴

EXPLANATION:

The expansion for the general term  (a + b)ⁿ is,

$\boxed{\bold{\large{\gray{T_{r+1}= \ ^nC_r\ a^{n-r}\ b^r}}}}$

8 in the power, also it is even

$\boxed{\bold{\large{\gray{Middle\ term = \dfrac{n + 2}{2}}}}}$

$\sf Middle\ term = \dfrac{8 + 2}{2}$

$\sf Middle\ term = \dfrac{10}{2}$

Middle term = 5

∴ 5th term is the middle term

r + 1 = 5

r = 4

n = 8

a = 7p

b = 2q

$\sf T_5= \ ^{8}C_4\ (7p)^{8-4}\ (2q)^4$

$\sf T_5= \ ^{8}C_4\ (7p)^{4}\ (2q)^4$

$\sf T_5= \ ^{8}C_4\ (7^{4}\ p^{4}\ 2^{4} q^4)$

$\sf T_5= \ ^{8}C_4\ 14^{4}\ (pq)^{4}$

$\boxed{\large{\sf {Middle \ term = \ ^{8}C_4\ 14^{4}\ (pq)^{4}}}}$

QUESTION 3:

Find the middle term of (4x² + 5x³)¹⁷.

ANSWER 3:

• ¹⁷C₉ (4)⁸ (5)⁹ (x)⁴³

EXPLANATION:

The expansion for the general term  (a + b)ⁿ is,

$\boxed{\bold{\large{\gray{T_{r+1}= \ ^nC_r\ a^{n-r}\ b^r}}}}$

17 in the power, also it is odd. So it has two middle terms

$\boxed{\bold{\large{\gray{Middle\ terms = \dfrac{n + 1}{2},\ \dfrac{n + 1}{2}+1}}}}$

$\sf Middle\ terms = \dfrac{17 + 1}{2},\ \dfrac{17 + 1}{2} +1$

$\sf Middle\ terms = \dfrac{18}{2},\ \dfrac{18}{2}+1$

Middle terms = 9, 10

∴ 9, 10th terms are the middle terms

r + 1 = 9, r + 1 = 10

r = 8, r = 9

n = 17

a = 4x²

b = 5x³

9th term :

$\sf T_9= \ ^{17}C_8\ (4x^2)^{17-8}\ (5x^3)^8$

$\sf T_9= \ ^{17}C_8\ (4x^2)^{9}\ (5x^3)^8$

$\sf T_9= \ ^{17}C_8\ (4^{9}x^{18})\ (5^8x^{24})$

$\sf T_9= \ ^{17}C_8\ 4^{9}\ 5^8\ x^{42}$

10th term :

$\sf T_{10}= \ ^{17}C_9\ (4x^2)^{17-9}\ (5x^3)^9$

$\sf T_{10}= \ ^{17}C_8\ (4x^2)^{8}\ (5x^3)^9$

$\sf T_{10}= \ ^{17}C_8\ (4^{8}x^{16})\ (5^9x^{27})$

$\sf T_9= \ ^{17}C_8\ 4^{8}\ 5^9\ x^{43}$

$\boxed{\large{\sf {Middle \ term= \ ^{17}C_8\ 4^{8}\ 5^9\ x^{43}}}}$

T₁₀ matches with the given option.

QUESTION 4 :

Find the middle term of (3/a³ - 5a⁴)²⁰.

ANSWER 4:

• ²⁰C₁₀ (15)¹⁰ a¹⁰.

EXPLANATION:

The expansion for the general term  (a + b)ⁿ is,

$\boxed{\bold{\large{\gray{T_{r+1}= \ ^nC_r\ a^{n-r}\ b^r}}}}$

20 in the power, also it is even

$\boxed{\bold{\large{\gray{Middle\ term = \dfrac{n + 2}{2}}}}}$

$\sf Middle\ term = \dfrac{20 + 2}{2}$

$\sf Middle\ term = \dfrac{22}{2}$

Middle term = 11

∴ 11th term is the middle term

r + 1 = 11

r = 10

n = 20

a = 3/a³

b =  5a⁴

$\sf T_{11}= \ ^{20}C_{10}\ \left(\dfrac{3}{a^3}\right)^{20-10}\ (5a^4)^{10}$

$\sf T_{11}= \ ^{20}C_{10}\ \left(\dfrac{3}{a^3}\right)^{10}\ (5^{10}a^{40})$

$\sf T_{11}= \ ^{20}C_{10}\ ({3}{a^{-3}})^{10}\ (5^{10}a^{40})$

$\sf T_{11}= \ ^{20}C_{10}\ ({3}^{10}{a^{-30}})\ (5^{10}a^{40})$

$\sf T_{11}= \ ^{20}C_{10}\ {15}^{10}\ a^{-30+40}$

$\sf T_{11}= \ ^{20}C_{10}\ {15}^{10}\ a^{10}$

$\boxed{\large{\sf {Middle \ term = \ ^{20}C_{10}\ {15}^{10}\ a^{10}}}}$

FINAL ANSWER:

• a - r

• b - p

• c - s

• d - q