In the previous section, we saw how to expand binomials by applying the general form of the binomial theorem. In this section, we will see some solved examples.
Solved example 8.1
Expand $\left(x^2+ \frac{3}{x} \right)^4, ~x \ne 0$ using binomial theorem
Solution:
1. We have: $(a+b)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;a^{n-k}\;b^k}$
2. In our present case, $a=x^2,~b=\frac{3}{x}~\rm{and}~n=4$
So we can write:
$\left(x^2+ \frac{3}{x} \right)^4~=~\sum\limits_{k\,=\,0}^{k\,=\,4}{{}^4 {\rm{C}}_k\;\left(x^2 \right)^{4-k}\;\left(\frac{3}{x} \right)^k}$
3. Thus we get:
$\left(x^2+ \frac{3}{x} \right)^4$
$\begin{array}{ll}
{}={}&{}^4 {\rm{C}}_0 \,\left(x^2 \right)^4\,\left(\frac{3}{x} \right)^0
&{}+{}& {}^4 {\rm{C}}_1\, \left(x^2 \right)^{4-1}\,\left(\frac{3}{x} \right)^1
&{}+{}& {}^4 {\rm{C}}_2\,\left(x^2 \right)^{4-2}\,\left(\frac{3}{x} \right)^2
&{}+{}& {}^4 {\rm{C}}_3\, \left(x^2 \right)^{4-3}\,\left(\frac{3}{x} \right)^3
&{}+{}& {}^4 {\rm{C}}_4\, \left(x^2 \right)^{4-4}\,\left(\frac{3}{x} \right)^4 \\
{}={}&1 × \left(x^2 \right)^4 × \left(\frac{3}{x} \right)^0
&{}+{}& 4 × \left(x^2 \right)^3 × \left(\frac{3}{x} \right)^1
&{}+{}& 6 × \left(x^2 \right)^2 × \left(\frac{3}{x} \right)^2
&{}+{}& 4 × \left(x^2 \right)^1 × \left(\frac{3}{x} \right)^3
&{}+{}& 1 × \left(x^2 \right)^0 × \left(\frac{3}{x} \right)^4 \\
{}={}&x^8
&{}+{}& 12 x^5
&{}+{}& 54 x^2
&{}+{}& \frac{108}{x}
&{}+{}& \frac{81}{x^4} \\
\end{array}$
Solved example 8.2
Compute 985
Solution:
1. First we write 98 as the sum or difference of two numbers.
• Those two numbers must be such that, their powers are easy to calculate.
• So we write: 98 = (100 - 2)
♦ Powers of 100 can be easily calculated.
♦ Powers of 2 can also be easily calculated.
2. Now we can write:
$98^5~=~(100-2)^5$
3. For expanding this, we can use the formula:
$(a-b)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;(-1)^k \, a^{n-k}\;b^k}$
4. Thus we get:
$(100-2)^5$
$\begin{array}{ll}
{}={}&{}^5 {\rm{C}}_0 \,100^5\,(-1)^0\,2^0
&{}+{}& {}^5 {\rm{C}}_1 × 100^{5-1} × (-1)^1 × 2^1
&{}+{}& {}^5 {\rm{C}}_2 × 100^{5-2} × (-1)^2 × 2^2
&{}+{}& {}^5 {\rm{C}}_3 × 100^{5-3} × (-1)^3 × 2^3
&{}+{}& {}^5 {\rm{C}}_4 × 100^{5-4} × (-1)^4 × 2^4
&{}+{}& {}^5 {\rm{C}}_5 × 100^{5-5} × (-1)^5 × 2^5&{}& {}
&{} \\
{}={}&1 × 10^{10} × 1 × 1
&{}+{}& 5 × 10^8 × -1 × 2
&{}+{}& 10 × 10^6 × 1 × 4
&{}+{}& 10 ×
10^4 × -1 × 8
&{}+{}& 5 ×
10^2 × 1 × 16
&{}+{}& 1 × 100^0 × -1 × 32 \\
{}={}&1 × 10^{10} × 1
&{}-{}& 5 × 10^8 × 2
&{}+{}& 10 × 10^6 × 4
&{}-{}& 10 ×
10^4 × 8
&{}+{}& 5 ×
10^2 × 16
&{}-{}& 1 × 100^0 × 32 \\
{}={}&9039207968 \\
\end{array}$
Solved example 8.3
Which is larger (1.01)1000000 or 10000 ?
Solution:
1. First we must find (1.01)1000000. For that, we write 1.01 as the sum or difference of two numbers.
• Those two numbers must be such that, their powers are easy to calculate.
• So we write: 1.01 = (1 + 0.01)
♦ Powers of 1 can be easily calculated.
♦ Powers of 0.01 can also be easily calculated.
2. Now we can write:
$(1.01)^{1000000}~=~(1+0.01)^{1000000}$
3. For expanding this, we can use the formula:
$(1+x)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;x^k}$
4. Thus we get:
$(1+0.01)^{1000000}$
$\begin{array}{ll}
{}={}&{}^{1000000} {\rm{C}}_0 × (0.01)^0
&{}+{}& {}^{1000000} {\rm{C}}_1 × (0.01)^1
&{}+{}& {}^{1000000} {\rm{C}}_2 × (0.01)^2
&{}+{}& {}^{1000000} {\rm{C}}_3 × (0.01)^3
&{}+{}& .~.~.
&{}+{}& {}^{1000000} {\rm{C}}_{1000000} × (0.01)^{1000000} \\
{}={}&1 × 1
&{}+{}& 1000000 × 0.01
&{}+{}& \text{[a +ve term]}
&{}+{}& \text{[a +ve term]}
&{}+{}& .~.~.
&{}+{}& \text{[a +ve term]} \\
{}={}&1
&{}+{}& 10000
&{}+{}& \text{[a +ve term]}
&{}+{}& \text{[a +ve term]}
&{}+{}& .~.~.
&{}+{}& \text{[a +ve term]} \\
{}={}&10001
&{}+{}& \text{[+ve terms]} \\
\end{array}$
5. 10001 + [+ve terms] will be greater than 10000
• So (1.01)1000000 is greater than 10000
6. Note that, all terms within the square brackets must be +ve terms. Otherwise there is no guarantee that 10001 + [+ve terms] will be greater than 10000
Solved example 8.4
Using binomial theorem, prove that 6n – 5n always leaves remainder 1 when divided by 25.
Solution:
1. Let a and b be two natural numbers.
• Suppose that, when a is divided by b, the quotient is q and remainder is r.
• Then we will be able to write: a = bq + r
♦ Where q and r are natural numbers.
2. In a similar way,
• If 6n – 5n, when divided by 25, gives quotient m and remainder 1, we will be able to write: 6n – 5n = 25m + 1, where m is a natural number.
• So our task is to prove that, 6n – 5n is equal to 25m + 1
3. We have the formula:
$(1+x)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;x^k}$
Put x = 5. Then we get:
$(1+5)^n~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{{}^n {\rm{C}}_k\;5^k}$
4. Thus we get:
$(1+5)^n~=~6^n$
$\begin{array}{ll}
{}={}&{}^n {\rm{C}}_0 × 5^0
&{}+{}& {}^n {\rm{C}}_1 × 5^1
&{}+{}& {}^n {\rm{C}}_2 × 5^2
&{}+{}& {}^n {\rm{C}}_3 × 5^3
&{}+{}& .~.~.
&{}+{}& {}^n {\rm{C}}_n × 5^4 \\
{}={}&1 × 1
&{}+{}& n × 5
&{}+{}& {}^n {\rm{C}}_2 × 5^2
&{}+{}& {}^n {\rm{C}}_3 × 5^3
&{}+{}& .~.~.
&{}+{}& {}^n {\rm{C}}_n × 5^4 \\
{}={}&1
&{}+{}& 5n
&{}+{}& {}^n {\rm{C}}_2 × 5^2
&{}+{}& {}^n {\rm{C}}_3 × 5^3
&{}+{}& .~.~.
&{}+{}& {}^n {\rm{C}}_n × 5^n \\
\end{array}$
5. We can write:
$6^n~=~1~+~5n~+~{}^n {\rm{C}}_2 × 5^2~+~{}^n {\rm{C}}_3 × 5^3
~+~.~.~.~+~{}^n {\rm{C}}_n × 5^n$
• This can be rearranged as:
$6^n-5n~=~1~+~{}^n {\rm{C}}_2 × 5^2~+~{}^n {\rm{C}}_3 × 5^3
~+~.~.~.~+~{}^n {\rm{C}}_n × 5^n$
⇒ $6^n-5n~=~1~+~5^2 \left[{}^n {\rm{C}}_2~+~{}^n {\rm{C}}_3 × 5^1
~+~.~.~.~+~{}^n {\rm{C}}_n × 5^{n-2}\right]$
6. $\left[{}^n {\rm{C}}_2~+~{}^n {\rm{C}}_3 × 5^1
~+~.~.~.~+~{}^n {\rm{C}}_n × 5^{n-2}\right]$ is a natural number m. So the result in (5) becomes:
6n - 5n = 1 + 52 × m
⇒ 6n - 5n = 25m + 1
Hence proved.
The link below gives some more solved examples.
In the next section we will see some solved examples.
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