1. Expand `(x+2)^4` using Binomial theorem method

Solution:
`(x+2)^4`

Using Binomial Theorem,

`(a+b)^n=((n),(0))a^(n)b^(0)+((n),(1))a^(n-1)b^(1)+((n),(2))a^(n-2)b^(2)+...+((n),(n))a^(0)b^(n)`

where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`

`"Here "a=x,b=2,n=4`

Now, `(x+2)^4`

`=((4),(0))(x)^4(2)^0+((4),(1))(x)^3(2)^1+((4),(2))(x)^2(2)^2+((4),(3))(x)^1(2)^3+((4),(4))(x)^0(2)^4`

`=1(x)^4(2)^0+4(x)^3(2)^1+(4!)/(2!(4-2)!)(x)^2(2)^2+(4!)/(3!(4-3)!)(x)^1(2)^3+1(x)^0(2)^4`

`=1(x^4)(1)+4(x^3)(2)+(4*3)/(2*1)(x^2)(4)+(4)/(1)(x)(8)+1(1)(16)`

`=1(x^4)(1)+4(x^3)(2)+6(x^2)(4)+4(x)(8)+1(1)(16)`

`=x^4+8x^3+24x^2+32x+16`

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2. Expand `(2x-3y)^4` using Binomial theorem method

Solution:
`(2x-3y)^4`

Using Binomial Theorem,

`(a+b)^n=((n),(0))a^(n)b^(0)+((n),(1))a^(n-1)b^(1)+((n),(2))a^(n-2)b^(2)+...+((n),(n))a^(0)b^(n)`

where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`

`"Here "a=2x,b=-3y,n=4`

Now, `(2x+(-3y))^4`

`=((4),(0))(2x)^4(-3y)^0+((4),(1))(2x)^3(-3y)^1+((4),(2))(2x)^2(-3y)^2+((4),(3))(2x)^1(-3y)^3+((4),(4))(2x)^0(-3y)^4`

`=1(2x)^4(-3y)^0+4(2x)^3(-3y)^1+(4!)/(2!(4-2)!)(2x)^2(-3y)^2+(4!)/(3!(4-3)!)(2x)^1(-3y)^3+1(2x)^0(-3y)^4`

`=1(16x^4)(1)+4(8x^3)(-3y)+(4*3)/(2*1)(4x^2)(9y^2)+(4)/(1)(2x)(-27y^3)+1(1)(81y^4)`

`=1(16x^4)(1)+4(8x^3)(-3y)+6(4x^2)(9y^2)+4(2x)(-27y^3)+1(1)(81y^4)`

`=16x^4-96x^3y+216x^2y^2-216xy^3+81y^4`

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3. Expand `(2x+5)^5` using Binomial theorem method

Solution:
`(2x+5)^5`

Using Binomial Theorem,

`(a+b)^n=((n),(0))a^(n)b^(0)+((n),(1))a^(n-1)b^(1)+((n),(2))a^(n-2)b^(2)+...+((n),(n))a^(0)b^(n)`

where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`

`"Here "a=2x,b=5,n=5`

Now, `(2x+5)^5`

`=((5),(0))(2x)^5(5)^0+((5),(1))(2x)^4(5)^1+((5),(2))(2x)^3(5)^2+((5),(3))(2x)^2(5)^3+((5),(4))(2x)^1(5)^4+((5),(5))(2x)^0(5)^5`

`=1(2x)^5(5)^0+5(2x)^4(5)^1+(5!)/(2!(5-2)!)(2x)^3(5)^2+(5!)/(3!(5-3)!)(2x)^2(5)^3+(5!)/(4!(5-4)!)(2x)^1(5)^4+1(2x)^0(5)^5`

`=1(32x^5)(1)+5(16x^4)(5)+(5*4)/(2*1)(8x^3)(25)+(5*4)/(2*1)(4x^2)(125)+(5)/(1)(2x)(625)+1(1)(3125)`

`=1(32x^5)(1)+5(16x^4)(5)+10(8x^3)(25)+10(4x^2)(125)+5(2x)(625)+1(1)(3125)`

`=32x^5+400x^4+2000x^3+5000x^2+6250x+3125`

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4. Expand `(x/2+1/x)^5` using Binomial theorem method

Solution:
`(x/2+1/x)^5`

Using Binomial Theorem,

`(a+b)^n=((n),(0))a^(n)b^(0)+((n),(1))a^(n-1)b^(1)+((n),(2))a^(n-2)b^(2)+...+((n),(n))a^(0)b^(n)`

where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`

`"Here "a=x/2,b=1/x,n=5`

Now, `(x/2+1/x)^5`

`=((5),(0))(x/2)^5(1/x)^0+((5),(1))(x/2)^4(1/x)^1+((5),(2))(x/2)^3(1/x)^2+((5),(3))(x/2)^2(1/x)^3+((5),(4))(x/2)^1(1/x)^4+((5),(5))(x/2)^0(1/x)^5`

`=1(x/2)^5(1/x)^0+5(x/2)^4(1/x)^1+(5!)/(2!(5-2)!)(x/2)^3(1/x)^2+(5!)/(3!(5-3)!)(x/2)^2(1/x)^3+(5!)/(4!(5-4)!)(x/2)^1(1/x)^4+1(x/2)^0(1/x)^5`

`=1((x^5)/32)(1)+5((x^4)/16)(1/x)+(5*4)/(2*1)((x^3)/8)(1/(x^2))+(5*4)/(2*1)((x^2)/4)(1/(x^3))+(5)/(1)(x/2)(1/(x^4))+1(1)(1/(x^5))`

`=1((x^5)/32)(1)+5((x^4)/16)(1/x)+10((x^3)/8)(1/(x^2))+10((x^2)/4)(1/(x^3))+5(x/2)(1/(x^4))+1(1)(1/(x^5))`

`=(x^5)/32+(5x^3)/16+(5x)/4+5/(2x)+5/(2x^3)+1/(x^5)`

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5. Expand `(101)^5` using Binomial theorem method

Solution:
`(101)^5`

Using Binomial Theorem,

`(a+b)^n=((n),(0))a^(n)b^(0)+((n),(1))a^(n-1)b^(1)+((n),(2))a^(n-2)b^(2)+...+((n),(n))a^(0)b^(n)`

where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`

`"Here "a=100,b=1,n=5`

Now, `(100+1)^5`

`=((5),(0))(100)^5(1)^0+((5),(1))(100)^4(1)^1+((5),(2))(100)^3(1)^2+((5),(3))(100)^2(1)^3+((5),(4))(100)^1(1)^4+((5),(5))(100)^0(1)^5`

`=1(100)^5(1)^0+5(100)^4(1)^1+(5!)/(2!(5-2)!)(100)^3(1)^2+(5!)/(3!(5-3)!)(100)^2(1)^3+(5!)/(4!(5-4)!)(100)^1(1)^4+1(100)^0(1)^5`

`=1(10000000000)(1)+5(100000000)(1)+(5*4)/(2*1)(1000000)(1)+(5*4)/(2*1)(10000)(1)+(5)/(1)(100)(1)+1(1)(1)`

`=1(10000000000)(1)+5(100000000)(1)+10(1000000)(1)+10(10000)(1)+5(100)(1)+1(1)(1)`

`=10000000000+500000000+10000000+100000+500+1`

`=10510100501`

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6. Expand `(98)^6` using Binomial theorem method

Solution:
`(98)^6`

Using Binomial Theorem,

`(a+b)^n=((n),(0))a^(n)b^(0)+((n),(1))a^(n-1)b^(1)+((n),(2))a^(n-2)b^(2)+...+((n),(n))a^(0)b^(n)`

where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`

`"Here "a=100,b=-2,n=6`

Now, `(100+(-2))^6`

`=((6),(0))(100)^6(-2)^0+((6),(1))(100)^5(-2)^1+((6),(2))(100)^4(-2)^2+((6),(3))(100)^3(-2)^3+((6),(4))(100)^2(-2)^4+((6),(5))(100)^1(-2)^5+((6),(6))(100)^0(-2)^6`

`=1(100)^6(-2)^0+6(100)^5(-2)^1+(6!)/(2!(6-2)!)(100)^4(-2)^2+(6!)/(3!(6-3)!)(100)^3(-2)^3+(6!)/(4!(6-4)!)(100)^2(-2)^4+(6!)/(5!(6-5)!)(100)^1(-2)^5+1(100)^0(-2)^6`

`=1(1000000000000)(1)+6(10000000000)(-2)+(6*5)/(2*1)(100000000)(4)+(6*5*4)/(3*2*1)(1000000)(-8)+(6*5)/(2*1)(10000)(16)+(6)/(1)(100)(-32)+1(1)(64)`

`=1(1000000000000)(1)+6(10000000000)(-2)+15(100000000)(4)+20(1000000)(-8)+15(10000)(16)+6(100)(-32)+1(1)(64)`

`=1000000000000-120000000000+6000000000-160000000+2400000-19200+64`

`=885842380864`

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7. Expand `(0.99)^5` using Binomial theorem method

Solution:
`(0.99)^5`

Using Binomial Theorem,

`(a+b)^n=((n),(0))a^(n)b^(0)+((n),(1))a^(n-1)b^(1)+((n),(2))a^(n-2)b^(2)+...+((n),(n))a^(0)b^(n)`

where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`

`"Here "a=1,b=-0.01,n=5`

Now, `(1+(-0.01))^5`

`=((5),(0))(1)^5(-0.01)^0+((5),(1))(1)^4(-0.01)^1+((5),(2))(1)^3(-0.01)^2+((5),(3))(1)^2(-0.01)^3+((5),(4))(1)^1(-0.01)^4+((5),(5))(1)^0(-0.01)^5`

`=1(1)^5(-0.01)^0+5(1)^4(-0.01)^1+(5!)/(2!(5-2)!)(1)^3(-0.01)^2+(5!)/(3!(5-3)!)(1)^2(-0.01)^3+(5!)/(4!(5-4)!)(1)^1(-0.01)^4+1(1)^0(-0.01)^5`

`=1(1)(1)+5(1)(-0.01)+(5*4)/(2*1)(1)(0.0001)+(5*4)/(2*1)(1)(-0.000001)+(5)/(1)(1)(0.00000001)+1(1)(-0.0000000001)`

`=1(1)(1)+5(1)(-0.01)+10(1)(0.0001)+10(1)(-0.000001)+5(1)(0.00000001)+1(1)(-0.0000000001)`

`=1-0.05+0.001-0.00001+0.00000005-0.0000000001`

`=0.9509900499`

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