# BINOMIAL THEOREM WORKSHEET

## About "Binomial theorem worksheet"

Binomial theorem worksheet :

Here we are going to see some practice questions using binomial theorem.

Binomial expansion for (x + a)n is,

nc0xna+ nc1xn-1a+ nc2xn-2a+ .........+ ncnxn-na0

## Binomial theorem worksheet - Questions

(1)  Find the expansion of (2x + 3y)5

(2)  Find the expansion of (3a + 5b)5

(3)  Find the expansion of (a - 2b)5

(4)  Find the expansion of (2x – 5y)7

## Binomial theorem worksheet - Solution

Question 1 :

Find the expansion of (2x + 3y)5

Solution :

Comparing the given question with (x + a)n

we get x  =  2x, a  = 3y and n  =  5

Since we have power 5, we are going to have 6 terms in the expansion

(2x+3y)5 = 5c0(2x)5(3y)+ 5c1(2x)4(3y)+ 5c2(2x)3(3y)+ 5c3(2x)2(3y)+ 5c4(2x)1(3y)4+ 5c5(2x)0(3y)5

While calculating the values of 5c0, 5c1 ,......... we can follow the given tricks

The value of nc0 and ncn is 1.

The value of nc1 is n.

The value of 5c2 means, in the numerator we should write 5 as 5  4 and in the denominator we should write 2  1

5c0  =  1, 5c1  =  5, 5c2  =  (54)/(21) ==> 20/2 = 10

5c3  =  (543)/(321)  ==> 60/6 = 10

5c4  =  (5432)/(4321)  ==>  5,  5c5 = 1

=  32x+ 5(16x4)(3y)+ 10(8x3)(9y) + 10(4x2)(27y3)  + 5(2x)(81y4) + 1(243 y5)

= 32x+240x4y + 720x3y2 + 1080x2y3  + 810xy4 + 243 y5

Question 2 :

Find the expansion of (3a + 5b)5

Solution :

Comparing the given question with (x + a)n

we get x  =  3a, a  = 5b and n  =  5

(3a + 5b)5

Since we have power 5, we are going to have 6 terms in the expansion

=  5c0(3a)5(5b)+ 5c1(3a)4(5b)+ 5c2(3a)3(5b)+ 5c3(3a)2(5b)+ 5c4(3a)1(5b)4+ 5c5(3a)0(5b)5

5c0  =  1, 5c1  =  5, 5c2  =  (54)/(21) ==> 20/2 = 10

5c3  =  (543)/(321)  ==> 60/6 = 10

5c4  =  (5432)/(4321)  ==>  5,  5c5 = 1

=  5c0(3a)5(5b)+ 5c1(3a)4(5b)+ 5c2(3a)3(5b)+ 5c3(3a)2(5b)+ 5c4(3a)1(5b)4+ 5c5(3a)0(5b)5

=  243a5 + 5(81a4)(5b) + 10 (27a3)(25b2) + 10(9a2)(125b3) + 5(3a) (625b4) + 1 (3125b5)

=  243a+ 2025a4b + 6750a3b+ 11250a2b+ 9375ab4+ 3125b5

Let us see the solution of next problem on "Binomial theorem worksheet".

Question 3 :

Find the expansion of (a - 2b)5

Solution :

Comparing the given question with (x + a)n

we get x  =  a, a  = -2b and n  =  5

(a - 2b)5

Since we have power 5, we are going to have 6 terms in the expansion

=  5c0(a)5(-2b)+ 5c1(a)4(-2b)+ 5c2(a)3(-2b)+ 5c3(a)2(-2b)+ 5c4(a)1(-2b)4+ 5c5(a)0(-2b)5

=  a- 10 ab + 40 ab2 - 80 ab+ 80 b4-32 b5

Question 4 :

Find the expansion of (2x – 5y)7

Solution :

Since we have power 7, we are going to have 8 terms in the expansion

(2x – 5y)7

= 7C0 (2x)7(–5y)0 + 7C1 (2x)6(–5y)1 + 7C2 (2x)5(–5y)2

+ 7C3 (2x)4(–5y)3 + 7C4 (2x)3(–5y)4 + 7C5 (2x)2(–5y)5

+ 7C6 (2x)1(–5y)6 + 7C7 (2x)0(–5y)7

7c0  =  1, 7c1  =  7, 7c2  =  (7⋅6)/(2⋅1) ==> 56/2 = 28

7c3  =  (7⋅6⋅5)/(3⋅2⋅1) ==> 210/6 = 35

7c4  =  (7⋅6⋅5⋅4)/(4⋅3⋅2⋅1) = 35

7c5  =  (7⋅6⋅5⋅4⋅3)/(5⋅4⋅3⋅2⋅1) = 21

7c6  =  (7⋅6⋅5⋅4⋅3⋅2)/(6⋅5⋅4⋅3⋅2⋅1) = 7

7c7  =  1

=  (1)(128x7)(1) + (7) (64x6) (–5y) + (21) (32x5) (25y2) + (35) (16x4) (–125y3) + (35) (8x3) (625y4) + (21) (4x2)(–3125y5) + (7) (2x) (15625y6) + (1)(1)(–78125y7)

= 128x7 – 2240x6y + 16800x5y2 – 70000x4y3 + 175000x3y4 – 262500x2y+ 218750xy6 – 78125y7 After having gone through the stuff given above, we hope that the students would have understood, "Binomial theorem worksheet".

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