Question

Using binomial theorem, evaluate \[{(102)^3}\]

Answer 1

Hint: If we have a small number like 2 or 5 we can find the cube root easily. But the given number is three digits so we use binomial expansion to solve this. We can write \[102 = 100 + 2\] that is \[{(102)^3} = {(100 + 2)^3}\] . It is in the form \[{(a + b)^n}\] . Now we can apply the binomial expansion. We know the combination formula \[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]

Complete step-by-step answer:
Now the binomial expansion of \[{(a + b)^n}\] is given by,
 \[{(a + b)^n}{ = ^n}{C_r}{a^n}{b^0}{ + ^n}{C_1}{a^{n - 1}}{b^1}{ + ^n}{C_2}{a^{n - 2}}{b^2} + .........{ + ^n}{C_n}{a^0}{b^n}\] ----- (1)
We first need to find \[{(a + b)^3}\] from above, then substituting the value of \[a\] and \[b\] we will get the required answer.
Now put \[n = 3\] in equation (1).
 \[ \Rightarrow {(a + b)^3}{ = ^3}{C_0}{a^3}{b^0}{ + ^3}{C_1}{a^{3 - 1}}{b^1}{ + ^3}{C_2}{a^{3 - 2}}{b^2}{ + ^3}{C_3}{a^{3 - 3}}{b^3}\]
 \[ \Rightarrow {(a + b)^3}{ = ^3}{C_0}{a^3}{b^0}{ + ^3}{C_1}{a^2}b{ + ^3}{C_2}a{b^2}{ + ^3}{C_3}{a^0}{b^3}\]
We know that any power of zero is one.
 \[ \Rightarrow {(a + b)^3}{ = ^3}{C_0}{a^3}{ + ^3}{C_1}{a^2}b{ + ^3}{C_2}a{b^2}{ + ^3}{C_3}{b^3}\] ----- (2)
Now, we need to find the combination values,
We have, \[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]
We know that \[0! = 1\] .
Now find the each combination values,
 \[{ \Rightarrow ^3}{C_0} = \dfrac{{3!}}{{0!(3 - 0)!}} = \dfrac{{3!}}{{3!}} = 1\] (Putting, \[n = 3,r = 0\] )
 \[{ \Rightarrow ^3}{C_1} = \dfrac{{3!}}{{1!(3 - 1)!}} = \dfrac{{3!}}{{2!}} = 3\] (Putting, \[n = 3,r = 1\] )
 \[{ \Rightarrow ^3}{C_2} = \dfrac{{3!}}{{2!(3 - 2)!}} = \dfrac{{3!}}{{2!}} = 3\] (Putting, \[n = 3,r = 2\] )
 \[{ \Rightarrow ^3}{C_3} = \dfrac{{3!}}{{3!(3 - 3)!}} = \dfrac{{3!}}{{3!}} = 1\] (Putting, \[n = 3,r = 3\] )
(All we did in above is simple calculation.)
Now, substituting these values in equation (2) we get,
 \[{(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\]
Now we have,
 \[{(102)^3} = {(100 + 2)^3}\]
 \[a = 100\] and \[b = 2\] , Substituting in above formula we get:
 \[{(100 + 2)^3} = {(100)^3} + \left( {3 \times {{(100)}^2} \times (2)} \right) + \left( {3 \times (100) \times {{(2)}^2}} \right) + {(2)^3}\]
Using simple multiplication, we get:
 \[{(102)^3} = 10,00,000 + \left( {6 \times 10000} \right) + \left( {3 \times 100 \times 4} \right) + 8\]
 \[{(102)^3} = 10,00,000 + 60000 + 1200 + 8\] (Adding all)
 \[{(102)^3} = 10,61,208\] is the required answer.
So, the correct answer is “10,61,208”.

Note: We can use the binomial expansion to find the values of cube root, fourth root, etc. easily. If they ask to find the value of \[{(212)^5}\] , express it as \[212 = 200 + 12\] and put \[n = 5\] . Follow the same procedure as we did above. Like this you can find any values easily. Careful when substituting the values and in the calculation part as we see in above it has a long calculation.