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Question

Question: Use a binomial theorem to find the value of \({(102)^3}\)....

Use a binomial theorem to find the value of (102)3{(102)^3}.

Explanation

Solution

Hint – In order to solve this problem you need to write 102 = 100+2. Then apply the binomial theorem to get the value of (102)3{(102)^3}.

Complete step-by-step answer:

We have to find the value of (102)3{(102)^3} using the binomial theorem.

We can write (102)3{(102)^3} as (100+2)3{(100 + 2)^3}.

We know that (a+b)n=nC0anb0+nC1an1b1+.......+nCna0bn{(a + b)^n} = {\,^n}{C_0}{a^n}{b^0} + {\,^n}{C_1}{a^{n - 1}}{b^1} + ....... + {\,^n}{C_n}{a^0}{b^n}. (Binomial Expansion)

So we can apply the same expansion in (100+2)3{(100 + 2)^3}.

So, (100+2)3=3C0100320+3C11003121+3C21003222+3C31003323 (100+2)3=3!0!(30)!(1000000)+3!1!(31)!(20000)+3!2!(32)!(400)+3!3!(33)!(8) (100+2)3=1000000+60000+1200+8 (100+2)3=1061208.  {(100 + 2)^3} = {\,^3}{C_0}{100^3}{2^0} + {\,^3}{C_1}{100^{^{3 - 1}}}{2^1} + {\,^3}{C_2}{100^{3 - 2}}{2^2} + {\,^3}{C_3}{100^{3 - 3}}{2^3} \\\ {(100 + 2)^3} = \dfrac{{3!}}{{0!(3 - 0)!}}(1000000) + \dfrac{{3!}}{{1!(3 - 1)!}}(20000) + \dfrac{{3!}}{{2!(3 - 2)!}}(400) + \dfrac{{3!}}{{3!(3 - 3)!}}(8) \\\ {(100 + 2)^3} = 1000000 + 60000 + 1200 + 8 \\\ {(100 + 2)^3} = 1061208. \\\
Hence, the answer to this question is 1061208.

Note – When you have asked to expand something with the help of binomial expansion then break it into two parts to apply the binomial expansion and then use the expansion (a+b)n=nC0anb0+nC1an1b1+.......+nCna0bn{(a + b)^n} = {\,^n}{C_0}{a^n}{b^0} + {\,^n}{C_1}{a^{n - 1}}{b^1} + ....... + {\,^n}{C_n}{a^0}{b^n} and solve to get the answer to this type of question.