📖 generic · CBSE Class 11 English medium · MATHEMATICS · Page 170question

BINOMIAL THEOREM · Part 2

Chapter 2: 4 P 2 = 4 C 2 ××××× 2! or ( · MATHEMATICS

all the rows of the Pascal’s triangle till index . This is a slightly lengthy process. The process, as you observe, will become more difficult, if we need the expansions involving still larger powers. We thus try to find a rule that will help us to find the expansion of the binomial for any power without writing all the rows of the Pascal’s triangle, that come before the row of the desired index.

For this, we make use of the concept of combinations studied earlier to rewrite the numbers in the Pascal’s triangle. We know that ! C !( )! r n – r , ≤ r ≤ n and n is a non-negative integer.

Also, n C = = n C n The Pascal’s triangle can now be rewritten as (Fig . ) Observing this pattern, we can now write the row of the Pascal’s triangle for any index without writing the earlier rows. For example, for the index the row would be C C C C C C C C . Thus, using this row and the observations (i), (ii) and (iii), we have ( a + b ) = C a + 7C a b + C a b + C a b + 7C a b + C a b + C ab + C b An expansion of a binomial to any positive integral index say n can now be visualised using these observations.

We are now in a position to write the expansion of a binomial to any positive integral index. Fig . Pascal’s triangle BINOMIAL THEOREM . .

Binomial theorem for any positive integer n , ( a + b ) n = n C a n + n C a n – b + n C a n – b + ...+

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