Question

1.2-10. Pascal's triangle gives a method for calculating the binomial coefficients: it begins as follows: 1464 1 15 10 10 5 The nth row of this triangle gives the coefficients for (a +b-. To find an entry in the table other than a on the boundary, add the two nearest numbers in the row directly above The equation 1I

Answer 1

Solution :-

working :-

Pascal's triangle decides the coefficients which emerge in binomial extensions. For a model, think about the development

$(x+y)^2=x^2+2xy+y^2$

$(x+y)^2=1*x^2*y^0+2*x^1y^1+1*x^0y^2$

Notice the coefficients are the numbers in line two of Pascal's triangle: 1, 2, 1. By and large, when a binomial like x + y is raised to a positive whole number power we have:

$(x + y)^n =a_0x^n + a_1x^n-1y + a_2x^n-2y^2 + ... + a_{n-1}xy^n-1 + a_ny^n$

where the coefficients ai in this development are absolutely the numbers on line n of Pascal's triangle. At the end of the day,

ai = {n choose i}.

This is the binomial hypothesis.

Notice that the whole right slanting of Pascal's triangle compares to the coefficient of $y^n$ in these binomial extensions, while the following corner to corner relates to the coefficient of $xy^{n-1}$, etc.

proof :-

Consider R.H.S, i.e   $\binom{n-1}{r}+\binom{n-1}{r-1}$

we can write it as,

$\binom{n-1}{r}+\binom{n-1}{r-1}=\frac{(n-1)!}{r!(n-r-1)!}+\frac{(n-1)!}{(r-1)!(n-r)!}$

$\binom{n-1}{r}+\binom{n-1}{r-1}=\frac{(n-1)!(n-r)+(n-1)!r}{r!(n-r)!}$

$\binom{n-1}{r}+\binom{n-1}{r-1}=\frac{((n-r)+r)(n-1)!}{r!(n-r)!}$

$\binom{n-1}{r}+\binom{n-1}{r-1}=\frac{n(n-1)!}{r!(n-r)!}$

$\binom{n-1}{r}+\binom{n-1}{r-1}=\frac{n!}{r!(n-r)!}$

$\binom{n-1}{r}+\binom{n-1}{r-1}=\binom{n}r{}$

Hence proved.