Binomial Theorem Solutions - Formula, Worksheet, Calculator and Examples.

Binomial Theorem Questions - Page 4
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However, for higher powers like 98^{2},101^{6}, etc., the calculations become difficult by using repeated multiplication. This difficulty was overcome by a theorem known as binomial theorem. It gives an easier way to expand , (a+b)^{n} where n is an integer o r a rational number.

Binomial Theorem for Positive Integral Indices

Let us have a look at the following identities done earlier

(a+b)^{0}=\: \: 1 a + b ≠ 0

(a+b)^{1}=\: \:a+b

(a+b)^{2}=\: \:a^{2}+2ab+b^{2}

(a+b)^{3}=\: \:a^{3}+3a^{2}b+3ab^{2}+b^{3}

(a+b)^{4}=\: \:(a+b)^{3}(a+b)\: =\: a^{4}4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4}

In these expansions, we observe that

  • The total number of terms in the expansion is one more than the index. For example, in the expansion of (a+b)^{2} , number of terms is 3 whereas the index of (a+b)^{2} , is 2
  • Powers of the first quantity ‘ a ’ go on decreasing by 1 whereas the powers of the second quantity ‘ b ’ increase by 1, in the successive terms.
  • In each term of the expansion, the sum of the indices of a and b is the same and is equal to the index of a + b
    We now arrange the coefficients in these expansions as follows (Fig8.1):

Do we observe any pattern in this table that will help us to write the next row? Yes we do. It can be seen that the addition of 1’s in the row for index 1 gives rise to 2 in the row for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each row. This can be continued till any index of our interest.

We can extend the pattern given in Fig 8.2 by writing a few more rows.

 

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