BINOMIAL THEOREM EXAMPLES

About "Binomial theorem examples"

Binomial theorem examples :

Here we are going to see some example problems on binomial theorem.

Binomial expansion for (x + a)n is,

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

Binomial theorem examples

Example 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

Let us see the next example on "Binomial theorem examples"

Example 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 next example on "Binomial theorem examples"

Example 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

Example 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

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