1. Expand upto `4^(th)` term of `sqrt(1-x)`

Solution:
`sqrt(1-x)`

Using Binomial Theorem,

`(1+a)^n=1+n*a+(n(n-1))/(2!)*a^2+(n(n-1)(n-2))/(3!)*a^3+...`

`"Here "a=-x,n=1/2`

Now, `(1-x)^(1/2)=1+(1/2)*-x+(1/2(1/2-1))/(2)*(-x)^2+(1/2(1/2-1)(1/2-2))/(6)*(-x)^3+...`

`=1+(1/2)*-x+(1/2((-1)/2))/(2)*(-x)^2+(1/2((-1)/2)((-3)/2))/(6)*(-x)^3+...`

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

`=1-x/2-(x^2)/8-(x^3)/16+...`

So, final answer is
`=1-x/2-(x^2)/8-(x^3)/16+...`

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2. Expand upto `4^(th)` term of `sqrt(2+x)`

Solution:
`sqrt(2+x)`

`=[2(1+x/2)]^(1/2)`

`=(2)^(1/2)(1+x/2)^(1/2)`

`=sqrt(2)(1+x/2)^(1/2)`

Using Binomial Theorem,

`(1+a)^n=1+n*a+(n(n-1))/(2!)*a^2+(n(n-1)(n-2))/(3!)*a^3+...`

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

Now, `(1+x/2)^(1/2)=1+(1/2)*x/2+(1/2(1/2-1))/(2)*(x/2)^2+(1/2(1/2-1)(1/2-2))/(6)*(x/2)^3+...`

`=1+(1/2)*x/2+(1/2((-1)/2))/(2)*(x/2)^2+(1/2((-1)/2)((-3)/2))/(6)*(x/2)^3+...`

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

`=1+x/4-(x^2)/32+(x^3)/128+...`

So, `sqrt(2)(1+x/2)^(1/2)`

`=sqrt(2)(1+x/4-(x^2)/32+(x^3)/128+...)`

`=sqrt(2)+(sqrt(2)x)/4-(sqrt(2)x^2)/32+(sqrt(2)x^3)/128+...`

So, final answer is
`=sqrt(2)+(sqrt(2)x)/4-(sqrt(2)x^2)/32+(sqrt(2)x^3)/128+...`

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3. Expand upto `4^(th)` term of `1/(1+5x)^(1/5)`

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

Using Binomial Theorem,

`(1+a)^n=1+n*a+(n(n-1))/(2!)*a^2+(n(n-1)(n-2))/(3!)*a^3+...`

`"Here "a=5x,n=(-1)/5`

Now, `(1+5x)^((-1)/5)=1+((-1)/5)*5x+((-1)/5((-1)/5-1))/(2)*(5x)^2+((-1)/5((-1)/5-1)((-1)/5-2))/(6)*(5x)^3+...`

`=1+((-1)/5)*5x+((-1)/5((-6)/5))/(2)*(5x)^2+((-1)/5((-6)/5)((-11)/5))/(6)*(5x)^3+...`

`=1-1/5*5x+3/25*(5x)^2-11/125*(5x)^3+...`

`=1-x+3x^2-11x^3+...`

So, final answer is
`=1-x+3x^2-11x^3+...`

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4. Expand upto `4^(th)` term of `sqrt(17)`

Solution:
`sqrt(17)`

`=(16+1)^(1/2)`

`=[16(1+1/16)]^(1/2)`

`=(16)^(1/2)(1+1/16)^(1/2)`

`=4(1+1/16)^(1/2)`

Using Binomial Theorem,

`(1+a)^n=1+n*a+(n(n-1))/(2!)*a^2+(n(n-1)(n-2))/(3!)*a^3+...`

`"Here "a=1/16,n=1/2`

Now, `(1+1/16)^(1/2)=1+(1/2)*1/16+(1/2(1/2-1))/(2)*(1/16)^2+(1/2(1/2-1)(1/2-2))/(6)*(1/16)^3+...`

`=1+(1/2)*1/16+(1/2((-1)/2))/(2)*(1/16)^2+(1/2((-1)/2)((-3)/2))/(6)*(1/16)^3+...`

`=1+1/2*1/16-1/8*(1/16)^2+1/16*(1/16)^3+...`

`=1+0.03125-0.000488+0.000015`

`=1.030777`

So, `4(1+1/16)^(1/2)`

`=4(1.030777)`

`=4.123108`

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5. Expand upto `4^(th)` term of `(27)^(1/2)`

Solution:
`(27)^(1/2)`

`=(25+2)^(1/2)`

`=[25(1+2/25)]^(1/2)`

`=(25)^(1/2)(1+2/25)^(1/2)`

`=5(1+2/25)^(1/2)`

Using Binomial Theorem,

`(1+a)^n=1+n*a+(n(n-1))/(2!)*a^2+(n(n-1)(n-2))/(3!)*a^3+...`

`"Here "a=2/25,n=1/2`

Now, `(1+2/25)^(1/2)=1+(1/2)*2/25+(1/2(1/2-1))/(2)*(2/25)^2+(1/2(1/2-1)(1/2-2))/(6)*(2/25)^3+...`

`=1+(1/2)*2/25+(1/2((-1)/2))/(2)*(2/25)^2+(1/2((-1)/2)((-3)/2))/(6)*(2/25)^3+...`

`=1+1/2*2/25-1/8*(2/25)^2+1/16*(2/25)^3+...`

`=1+0.04-0.0008+0.000032`

`=1.039232`

So, `5(1+2/25)^(1/2)`

`=5(1.039232)`

`=5.19616`

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