$(a+b)^6={ }_6 C_0 a^6+{ }_6 C_1 a^5 b+{ }_6 C_2 a^4 b^2+{ }_6 C_3 a^3 b^3+{ }_6 C_4 a^2 b^4+{ }_6 C_5 a b^5+{ }_6 C_6 b^6$



Evaluate $1.3^6$ using the Binomial Theorem.



Write 1.3 as a binomial: (1 + □)

Verified Answer :

  • Express \(1.3\) as a binomial sum \((1 + 0.3)\).
  • Apply the Binomial Theorem expansion for \((1 + b)^6\) using binomial coefficients \(\{1, 6, 15, 20, 15, 6, 1\}\).
  • Calculate the seven individual terms: \(1, 1.8, 1.35, 0.54, 0.1215, 0.01458,\) and \(0.000729\).
  • Sum the terms to arrive at the final result: \(\boxed{4.826809}\).

Explanation

  1. Rewriting 1.3 as a Binomial Expression

To evaluate \((1.3)^6\) using the Binomial Theorem, we first rewrite \(1.3\) as a binomial sum in the form \((a + b)\). Setting \(a = 1\) and \(b = 0.3\), we can express \(1.3\) as \((1 + 0.3)\). The Binomial Theorem states that:
\[(a+b)^6 = \binom{6}{0} a^6 b^0 + \binom{6}{1} a^5 b^1 + \binom{6}{2} a^4 b^2 + \binom{6}{3} a^3 b^3 + \binom{6}{4} a^2 b^4 + \binom{6}{5} a^1 b^5 + \binom{6}{6} a^0 b^6.\]
Substituting \(a=1\) and \(b=0.3\), the expression becomes:
\[(1 + 0.3)^6 = \binom{6}{0} (1)^6 (0.3)^0 + \binom{6}{1} (1)^5 (0.3)^1 + \binom{6}{2} (1)^4 (0.3)^2 + \binom{6}{3} (1)^3 (0.3)^3 + \binom{6}{4} (1)^2 (0.3)^4 + \binom{6}{5} (1)^1 (0.3)^5 + \binom{6}{6} (1)^0 (0.3)^6.\]

  1. Calculating Individual Terms of the Expansion

We recall the binomial coefficients for the 6th row of Pascal's Triangle: \(\binom{6}{0} = 1, \binom{6}{1} = 6, \binom{6}{2} = 15, \binom{6}{3} = 20, \binom{6}{4} = 15, \binom{6}{5} = 6, \binom{6}{6} = 1\).
Now, we calculate each term individually:

  1. \(1 \cdot 1^6 \cdot (0.3)^0 = 1 \cdot 1 \cdot 1 = 1\)
  2. \(6 \cdot 1^5 \cdot (0.3)^1 = 6 \cdot 1 \cdot 0.3 = 1.8\)
  3. \(15 \cdot 1^4 \cdot (0.3)^2 = 15 \cdot 1 \cdot 0.09 = 1.35\)
  4. \(20 \cdot 1^3 \cdot (0.3)^3 = 20 \cdot 1 \cdot 0.027 = 0.54\)
  5. \(15 \cdot 1^2 \cdot (0.3)^4 = 15 \cdot 1 \cdot 0.0081 = 0.1215\)
  6. \(6 \cdot 1^1 \cdot (0.3)^5 = 6 \cdot 1 \cdot 0.00243 = 0.01458\)
  7. \(1 \cdot 1^0 \cdot (0.3)^6 = 1 \cdot 1 \cdot 0.000729 = 0.000729\)
  1. Summing the Terms for Final Result

Summing all the calculated terms gives:
\[1 + 1.8 + 1.35 + 0.54 + 0.1215 + 0.01458 + 0.000729 = 4.826809.\]
Thus, the expansion confirms that \((1.3)^6 = \boxed{4.826809}\). The missing value in the box was \(0.3\).

Examples

The Binomial Theorem is a powerful tool in finance and statistics. For instance, if you are calculating compound interest where an investment grows by 30% each year for 6 years, the factor is \((1 + 0.3)^6\). The Binomial Theorem allows for a methodical way to compute this value without needing complex exponentiation, helping analysts predict the growth of assets over time.