Sunday, 7 November 2021

Binomial Theorem

EXERCISE -1

1) EXPAND:

a) (3x+2y)⁴.                81x⁴+216x³y+ 216x²y²+ 96xy³+ 16y⁴

b) (2x-3y)⁴.      16x⁴-96x³y+ 216x²y³- 216xy²+ 81y⁴

c) (x+ 1/y)¹¹.             x¹¹+ 11x¹⁰/y + 55x⁹/y² + 165x⁸/y³ + 330x⁷/y⁴ + 462x⁶/y⁵ + 462x⁵/y⁶ + 330x⁴/y⁷ + 165x³/y⁸ + 55x²/y⁹ + 11x/y¹⁰ + 1/y¹¹

d) (√x + √y)¹⁰.      x⁵+ 10x⁹⁾²y¹⁾²+ 45x⁴y+ 120x⁷⁾²y³⁾² + 210x³y²+252⁵⁾² y⁵⁾² + 210x²y³ + 120x³⁾²y⁷⁾²+ 45xy⁴ + 10x¹⁾²y⁹⁾² + y⁵

e) (x + 1/x)⁵.      x⁵+5x³+ 10x+ 10/x+ 5/x³+ 1/x⁵

f) (x - 1/x)⁶.              x⁶-6x⁴+15x²- 20+ 15/x²- 6/x⁴+ 1/x⁶

g) (2x/3 - 3/2x)⁶.             64x⁶/729 - 32x⁴/27 + 20x²/3 - 20+ 135/4x² - 243/8x⁴+ 729/64x⁶

h) (x² - 2/x)⁷.           x¹⁴ -14x¹¹ + 84x⁸ - 280x⁵ + 560x² - 672/x + 448/x⁴ - 128/x⁷

I) (1+x)⁵.        1+ 5x + 10x²+ 10x³+ 5x⁴ + x⁵

j) (1- 3x)⁷.       1- 21x+ 189x² - 945x³ + 2835x⁴ - 5103x⁵ + 5103x⁶- 2187x⁷

k)  (a² - 2bc)⁵.               a¹⁰-10a⁸bc+ 40a⁶b²c²- 80a⁴b³c³+80a²b⁴c⁴-32b⁵c⁵

L) (x³ + 2/x²)⁵.       x¹⁵+ 10x¹⁰+ 40x⁵ + 80+ 80/x⁵ + 32/x¹⁰

m) (1+ 2x- 3x²)⁵.       1+ 10x+ 25x²- 40x³ - 190x⁴+ 92x⁵+ 570x⁶- 360x⁷ - 675x⁸ + 810x⁹ - 243x¹⁰


2) EVALUATE::
a) (√2+1)⁶+(√2-1)⁶.                      198

b) (2+√3)⁷+(2 - √3)⁷.                 9884

c) (√3+1)⁵ +(√3 - 1)⁵                88√3

d) (x+ √(x²-1)⁶+ (x- √(x²-1)⁶.   64x⁶ - 96x⁴ + 36x² - 2.

e) (2a+b)⁶ - 6b(2a+b)⁵+ 15b²(2a+b)⁴ - 20b³(2a+b)³+15b⁴(2a+b)²- 6b⁵(2a+b) + b⁶.                         64a⁶

f) (1+2√x)⁵+(1-2√x)⁵.    2(1+40x+80x²)

g) (3+√2)⁵-(3-√2)⁵.                1178√2

h) (√3+1)⁵-(√3-1)⁵.                  152 

I) (96)³.                                  884736

j) (102)⁵.                    11040808032

k) (101)⁴.                         104060401

l) (98)⁵.                          9039207968

m) (994)⁴                   976215137296

n) (1.01)⁵. 1.0510100501

o) (1001)⁵.        1005010010005001


3) FIND:
a) 7th term of (4x/5 + 5/2x)⁸.    4375/x⁴

b) 10th term of (a/b - 2b/a²)¹².    -366080b⁵/a¹⁴

c) 16th term of (√x - √y)¹⁷.      -136xy¹⁵⁾²

d) 4th term of (x/y - y/x)¹⁰.   -120x⁴/y⁴

e) 5th term of (x²+ 3/x³)¹⁰.         17010

f) 19th term of (2√x - √y)²⁰.   760xy⁹

g) 11th term of (2x - 1/x²)²⁵.   ²⁵C₁₀(2¹⁵/x⁵)

h) 7th term in (3x² - 1/x³)¹⁰.     17010/x¹⁰

I) 7th term in (4x/5 + 5/2x)⁸.  4375/x⁴

j) 4th term in (x+ 2/x)⁹.          672x³


4) Find the coefficient of:
a) x⁷ in (x²+ 1/x)¹¹.                    462

b) x² in (3x- 1/x)⁶.                   1215

c) x¹⁸ in (x² - 3a/x)¹⁵.       110565a⁴

d)  x¹⁰ in (x² - 2)¹¹.                 29568

e) x⁶ in (3x² - 1/3x)⁹.                 378

f) x¹⁰ in (2x² -1/x)²⁰.          ²⁰C₁₀ 2¹⁰ 

g) x⁷ in (x - 1/x²).               -⁴⁰C₁₁

h) 1/x¹⁵ in (3x² - a/3x³)¹⁰.   -40a⁷/27

I) x⁹ in (x² - 1/3x)⁹.                  -28/9

j) 1/x² in (2x³ - 1/x²)⁶.               60

k) x¹² in (ax⁴ - bx)⁹.                9ab⁸

l) x³² in (x⁴ - 1/x³)¹⁵.                 1365

m) 1/x in (2x² - 1/x)¹⁰.             -960

n) 1/x¹¹ in (x² - 1/x³)¹².             -792


5) Find the term and coefficient independent of x in the expansion of::
a) (x+1/x)¹⁰.                              252

b) (x² +1/x)¹².                            495

c) (2x+1/3x²)⁹.                        1792/9

d) (x - 1/x)¹².                                924

e) (x² - 2/x³)¹⁵.                       320320

f) (3x²/2 -1/3x)⁹.                       7/18

g) (9x²-1/3x)¹².                          495

h) (√x - √c/√x)¹⁰.                -252√c⁵

I) (2x²- 3/x³)²⁵.            ²⁵C₁₀(2¹⁵× 3¹⁰)

j) {√(x/3)+3/2x²)¹⁰.      -3003×3¹⁰×2⁵



6) Find the middle term in the expansion of:

a) (3 + x)⁶.                               540x³                       
b) (2x/3 - 3/2x)²⁰.                 ²⁰C₁₀

c) (1 - x²/2)¹⁴.                -429x¹⁴/16

d) (a/x + bx)¹².                       924a⁶b⁶

e) (x² - 2/x)¹⁰.                     -8064x⁵

f) (a/3 + 9b)⁸.                    5670a⁴b⁴

g) (x/a - a/x)¹⁰.                         -252

h) (3x/ - x³/6)⁹.   189x¹⁷/8, -21x¹⁹/16

i) (2x² - 1/x)⁷.               -560x⁵, 280x²

j) (3x - 2/x²)¹⁵.         -6435×3⁸×2⁷/x⁶, 6437×3⁷×2⁸/x⁹

k) (x⁴ - 1/x³)¹¹.              -462x⁹, 462x²

I) (x - 1/x)¹⁰.                                -252

m) (2x - x²/4)⁹.     63x¹³/4, -63x¹⁴/32


7) a) Find the 5th term from the end in the expansion of (x - 1/x)¹².    495x⁴

b) Find the 4th term from the end in the expansion of (4x/5 - 5/2x)⁹.    10500/x³

c) (x⁴ + 1/x³)¹⁵, 4th term from the end.                       455x³⁹ or 455/x²⁴

d) (3/x² - x³/6)³, 4th term from the end.                   35x⁶/48

e) (2x - 1/x²)²⁵, 11th term from the end.                  - ²⁵C₁₅ 2¹⁰/x ²⁰

f) 11th term from end (2x - 1/x²)²⁵.     ²⁵C₁₅ (2¹⁰/x²⁰)

g) 5th term from the end of (3x -1/x²)¹⁰.                      17010/x⁸

h) 4th term from the end in (x+2/x)⁹.                 5376/x³

I) 4th term from the end of (4x/5 - 5/2x)⁹.                        10500/x³

j) 7th term from the end in (2x² - 3/2x)⁸.                      4032x¹⁰


EXERCISE -2

1) Using Binomial theorem, prove
A) 6ⁿ - 5n leaves remainder 1 when divided by 25.

B) 2³ⁿ - 7n -1 is divisible by 49, where n belongs to N.

C) 3ⁿ⁺² - 8n - 9 is divisible by 64, where n belongs to N.

D) 3³ⁿ - 26n -1 is divisible by 676.

E) 9ⁿ⁺¹ - 8n - 9 is divisible by 64, where n belongs to N.

2) Show that the middle term in the expansion of
A) (1+ x)²ⁿ is :
{1.3.5....(2n -1) 2ⁿ. xⁿ}/n!

B) (x+ 1/x)²ⁿ is :
{1.3.5....(2n -1) 2ⁿ}/n!

3) Find the general term in the expansion
a) (x² - y)⁶.           

4) Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion (⁴√2 + 1/⁴√3)ⁿ is √6: 1.                        10

5) If 17th and 18th term in the expansion of (2+ a)⁵⁰ are equal, then find a.

6) If the fourth term in the expansion of (ax + 1/x))ⁿ is 5/2, then find the value of a and n.          1/2, 6

7) Find the value of a so that the term independent of x in (√x + a/x²)¹⁰ is 405.                       ±3

8) Find a positive value of n for which the Coefficient of x² in the expansion of (1+ x)ⁿ is 6.              4

9) Find the coefficient of x⁷ in (ax² + 1/bx)¹¹ and 1/x⁷ in (ax - 1/bx²)¹¹.  11C5 a⁶/b⁵

10)A) Prove that the coefficient of (1+ x)²ⁿ is equal to the sum of the coefficient of middle terms in the expansion of (1+ x)²ⁿ⁻¹.
B) Find the value of k for which the coefficient of the middle term in (1+ Kx)⁴ and (1- Kx)⁶ are equal.    -3/10

11) The sum of the coefficient of first three terms in the expansion of (x - 3/x²)ⁿ, x≠ 0, n being a natural number, is 559. Find the term of the expansion containing x³.        12

12) If the coefficient of (2r+4)th and (r -2)th terms in the expansion of (1+ x)¹⁸ are equal, find r.             6

13) If the coefficient of (2r+1)th and (r +2)th terms in the expansion of (1+ x)⁴³ are equal, find r.            14

14) If the coefficient of (r - 5)th and (2r -1)th terms in the expansion of (1+ x)³⁴ are equal, find r.            14

15) The co-efficients of 5th, 6th, 7th terms in the expansion (1+ x)ⁿ are in AP, find n.                    7 or 14

16) The co-efficients of 2nd, 3rd, 4th terms in the expansion (1+ x)²ⁿ are in AP, show that 2n² - 9n +7= 0.

17) If the coefficient of (r - 1)th, r th and (r +1)th terms in the expansion of (1+ x)ⁿ are in the ratio 1:3:5 then find n, r.                                    7, 3

18) The 3rd, 4th, and 5th terms in the expansion of (x + a)ⁿ are respectively 84, 280 and 560, find the value of x, a, n.          1, 7, 2

19) If the coefficient of three consecutive terms in the expansion of (1+ x)ⁿ be 76, 95 and 76, find n. 8






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