Here `1-(1+x)^-120=94.7643x`

`:.-94.7643x+1-1/((1+x)^120)=0`

Let `f(x) = -94.7643x+1-1/((1+x)^120)`

`d/(dx)(-94.7643x+1-1/((1+x)^120))=-94.7643+120/((1+x)^121)`

`d/(dx)(-94.7643x+1-1/((1+x)^120))`

`=-d/(dx)(94.7643x)+d/(dx)(1)-d/(dx)(1/((1+x)^120))`

`d/(dx)(1/((1+x)^120))=-120/((1+x)^121)`

`d/(dx)(1/((1+x)^120))`

`=-120/((1+x)^121)*d/(dx)(1+x)`

`d/(dx)(1+x)=1`

`d/(dx)(1+x)`

`=d/(dx)(1)+d/(dx)(x)`

`=0+1`

`=1`

`=-120/((1+x)^121)*1`

`=-120/((1+x)^121)`

`=-94.7643+0-(-120/((1+x)^121))`

`=-94.7643+0+120/((1+x)^121)`

`=-94.7643+120/((1+x)^121)`

`:. f'(x) = -94.7643+120/((1+x)^121)`

`x_0 = 0.1`


`1^(st)` iteration :

`f(x_0)=f(0.1)=-94.7643*0.1+1-1/((1+0.1)^120)=-8.476441`

`f'(x_0)=f'(0.1)=-94.7643+120/((1+0.1)^121)=-94.763123`

`x_1 = x_0 - f(x_0)/(f'(x_0))`

`x_1=0.1 - (-8.476441)/(-94.763123)`

`x_1=0.010551`


`2^(nd)` iteration :

`f(x_1)=f(0.010551)=-94.7643*0.010551+1-1/((1+0.010551)^120)=-0.283674`

`f'(x_1)=f'(0.010551)=-94.7643+120/((1+0.010551)^121)=-61.06503`

`x_2 = x_1 - f(x_1)/(f'(x_1))`

`x_2=0.010551 - (-0.283674)/(-61.06503)`

`x_2=0.005906`


`3^(rd)` iteration :

`f(x_2)=f(0.005906)=-94.7643*0.005906+1-1/((1+0.005906)^120)=-0.052973`

`f'(x_2)=f'(0.005906)=-94.7643+120/((1+0.005906)^121)=-35.914489`

`x_3 = x_2 - f(x_2)/(f'(x_2))`

`x_3=0.005906 - (-0.052973)/(-35.914489)`

`x_3=0.004431`


`4^(th)` iteration :

`f(x_3)=f(0.004431)=-94.7643*0.004431+1-1/((1+0.004431)^120)=-0.008181`

`f'(x_3)=f'(0.004431)=-94.7643+120/((1+0.004431)^121)=-24.480398`

`x_4 = x_3 - f(x_3)/(f'(x_3))`

`x_4=0.004431 - (-0.008181)/(-24.480398)`

`x_4=0.004097`


`5^(th)` iteration :

`f(x_4)=f(0.004097)=-94.7643*0.004097+1-1/((1+0.004097)^120)=-0.000479`

`f'(x_4)=f'(0.004097)=-94.7643+120/((1+0.004097)^121)=-21.592676`

`x_5 = x_4 - f(x_4)/(f'(x_4))`

`x_5=0.004097 - (-0.000479)/(-21.592676)`

`x_5=0.004074`


Approximate root of the equation `-94.7643x+1-1/((1+x)^120)=0` using Newton Raphson method is `0.004074` (After 5 iterations)

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	0.1	-8.476441	-94.763123	0.010551	`x_0 = x_1`
2	0.010551	-0.283674	-61.06503	0.005906	`x_0 = x_1`
3	0.005906	-0.052973	-35.914489	0.004431	`x_0 = x_1`
4	0.004431	-0.008181	-24.480398	0.004097	`x_0 = x_1`
5	0.004097	-0.000479	-21.592676	0.004074	`x_0 = x_1`

Here `1-(1+x)^-60=42.0345x`

`:.-42.0345x+1-1/((1+x)^60)=0`

Let `f(x) = -42.0345x+1-1/((1+x)^60)`

`d/(dx)(-42.0345x+1-1/((1+x)^60))=-42.0345+60/((1+x)^61)`

`d/(dx)(-42.0345x+1-1/((1+x)^60))`

`=-d/(dx)(42.0345x)+d/(dx)(1)-d/(dx)(1/((1+x)^60))`

`d/(dx)(1/((1+x)^60))=-60/((1+x)^61)`

`d/(dx)(1/((1+x)^60))`

`=-60/((1+x)^61)*d/(dx)(1+x)`

`d/(dx)(1+x)=1`

`d/(dx)(1+x)`

`=d/(dx)(1)+d/(dx)(x)`

`=0+1`

`=1`

`=-60/((1+x)^61)*1`

`=-60/((1+x)^61)`

`=-42.0345+0-(-60/((1+x)^61))`

`=-42.0345+0+60/((1+x)^61)`

`=-42.0345+60/((1+x)^61)`

`:. f'(x) = -42.0345+60/((1+x)^61)`

`x_0 = 0.1`


`1^(st)` iteration :

`f(x_0)=f(0.1)=-42.0345*0.1+1-1/((1+0.1)^60)=-3.206734`

`f'(x_0)=f'(0.1)=-42.0345+60/((1+0.1)^61)=-41.855358`

`x_1 = x_0 - f(x_0)/(f'(x_0))`

`x_1=0.1 - (-3.206734)/(-41.855358)`

`x_1=0.023385`


`2^(nd)` iteration :

`f(x_1)=f(0.023385)=-42.0345*0.023385+1-1/((1+0.023385)^60)=-0.232823`

`f'(x_1)=f'(0.023385)=-42.0345+60/((1+0.023385)^61)=-27.387094`

`x_2 = x_1 - f(x_1)/(f'(x_1))`

`x_2=0.023385 - (-0.232823)/(-27.387094)`

`x_2=0.014884`


`3^(rd)` iteration :

`f(x_2)=f(0.014884)=-42.0345*0.014884+1-1/((1+0.014884)^60)=-0.037756`

`f'(x_2)=f'(0.014884)=-42.0345+60/((1+0.014884)^61)=-17.670589`

`x_3 = x_2 - f(x_2)/(f'(x_2))`

`x_3=0.014884 - (-0.037756)/(-17.670589)`

`x_3=0.012747`


`4^(th)` iteration :

`f(x_3)=f(0.012747)=-42.0345*0.012747+1-1/((1+0.012747)^60)=-0.003493`

`f'(x_3)=f'(0.012747)=-42.0345+60/((1+0.012747)^61)=-14.32809`

`x_4 = x_3 - f(x_3)/(f'(x_3))`

`x_4=0.012747 - (-0.003493)/(-14.32809)`

`x_4=0.012504`


`5^(th)` iteration :

`f(x_4)=f(0.012504)=-42.0345*0.012504+1-1/((1+0.012504)^60)=-0.00005`

`f'(x_4)=f'(0.012504)=-42.0345+60/((1+0.012504)^61)=-13.918183`

`x_5 = x_4 - f(x_4)/(f'(x_4))`

`x_5=0.012504 - (-0.00005)/(-13.918183)`

`x_5=0.0125`


Approximate root of the equation `-42.0345x+1-1/((1+x)^60)=0` using Newton Raphson method is `0.0125` (After 5 iterations)

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	0.1	-3.206734	-41.855358	0.023385	`x_0 = x_1`
2	0.023385	-0.232823	-27.387094	0.014884	`x_0 = x_1`
3	0.014884	-0.037756	-17.670589	0.012747	`x_0 = x_1`
4	0.012747	-0.003493	-14.32809	0.012504	`x_0 = x_1`
5	0.012504	-0.00005	-13.918183	0.0125	`x_0 = x_1`

Here `1-(1+x)^-48=39.4283x`

`:.-39.4283x+1-1/((1+x)^48)=0`

Let `f(x) = -39.4283x+1-1/((1+x)^48)`

`d/(dx)(-39.4283x+1-1/((1+x)^48))=-39.4283+48/((1+x)^49)`

`d/(dx)(-39.4283x+1-1/((1+x)^48))`

`=-d/(dx)(39.4283x)+d/(dx)(1)-d/(dx)(1/((1+x)^48))`

`d/(dx)(1/((1+x)^48))=-48/((1+x)^49)`

`d/(dx)(1/((1+x)^48))`

`=-48/((1+x)^49)*d/(dx)(1+x)`

`d/(dx)(1+x)=1`

`d/(dx)(1+x)`

`=d/(dx)(1)+d/(dx)(x)`

`=0+1`

`=1`

`=-48/((1+x)^49)*1`

`=-48/((1+x)^49)`

`=-39.4283+0-(-48/((1+x)^49))`

`=-39.4283+0+48/((1+x)^49)`

`=-39.4283+48/((1+x)^49)`

`:. f'(x) = -39.4283+48/((1+x)^49)`

`x_0 = 0.1`


`1^(st)` iteration :

`f(x_0)=f(0.1)=-39.4283*0.1+1-1/((1+0.1)^48)=-2.953137`

`f'(x_0)=f'(0.1)=-39.4283+48/((1+0.1)^49)=-38.97852`

`x_1 = x_0 - f(x_0)/(f'(x_0))`

`x_1=0.1 - (-2.953137)/(-38.97852)`

`x_1=0.024237`


`2^(nd)` iteration :

`f(x_1)=f(0.024237)=-39.4283*0.024237+1-1/((1+0.024237)^48)=-0.272413`

`f'(x_1)=f'(0.024237)=-39.4283+48/((1+0.024237)^49)=-24.581843`

`x_2 = x_1 - f(x_1)/(f'(x_1))`

`x_2=0.024237 - (-0.272413)/(-24.581843)`

`x_2=0.013155`


`3^(rd)` iteration :

`f(x_2)=f(0.013155)=-39.4283*0.013155+1-1/((1+0.013155)^48)=-0.052697`

`f'(x_2)=f'(0.013155)=-39.4283+48/((1+0.013155)^49)=-14.128063`

`x_3 = x_2 - f(x_2)/(f'(x_2))`

`x_3=0.013155 - (-0.052697)/(-14.128063)`

`x_3=0.009425`


`4^(th)` iteration :

`f(x_3)=f(0.009425)=-39.4283*0.009425+1-1/((1+0.009425)^48)=-0.00906`

`f'(x_3)=f'(0.009425)=-39.4283+48/((1+0.009425)^49)=-9.116324`

`x_4 = x_3 - f(x_3)/(f'(x_3))`

`x_4=0.009425 - (-0.00906)/(-9.116324)`

`x_4=0.008431`


`5^(th)` iteration :

`f(x_4)=f(0.008431)=-39.4283*0.008431+1-1/((1+0.008431)^48)=-0.000739`

`f'(x_4)=f'(0.008431)=-39.4283+48/((1+0.008431)^49)=-7.617439`

`x_5 = x_4 - f(x_4)/(f'(x_4))`

`x_5=0.008431 - (-0.000739)/(-7.617439)`

`x_5=0.008334`


`6^(th)` iteration :

`f(x_5)=f(0.008334)=-39.4283*0.008334+1-1/((1+0.008334)^48)=-0.000007`

`f'(x_5)=f'(0.008334)=-39.4283+48/((1+0.008334)^49)=-7.467189`

`x_6 = x_5 - f(x_5)/(f'(x_5))`

`x_6=0.008334 - (-0.000007)/(-7.467189)`

`x_6=0.008333`


Approximate root of the equation `-39.4283x+1-1/((1+x)^48)=0` using Newton Raphson method is `0.008333` (After 6 iterations)

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	0.1	-2.953137	-38.97852	0.024237	`x_0 = x_1`
2	0.024237	-0.272413	-24.581843	0.013155	`x_0 = x_1`
3	0.013155	-0.052697	-14.128063	0.009425	`x_0 = x_1`
4	0.009425	-0.00906	-9.116324	0.008431	`x_0 = x_1`
5	0.008431	-0.000739	-7.617439	0.008334	`x_0 = x_1`
6	0.008334	-0.000007	-7.467189	0.008333	`x_0 = x_1`