Unit-I: Sets and Functions
Unit-II: Algebra
Unit-III: Coordinate Geometry
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability

Pascal’s triangle

The Coefficient with binomial expansion can be arranged in the following triangle pattern named after Blaise Pascal.

\to (a-b)^n= \{a+(-b)\}^n

=a^n-^nC_1a^{n-1}b+^nC_2a^{n-2}b^2+...+(-1)^{rn}C_ra^{n-r}b^r+...+(-1)^{n-1} ^nC_{n-1}ab^{n-1}+(-1)^nb^n

 \to (1+x)^n=1+^nC_1x+^nC_2x^2+...+^nC_{n-1}x^{n-1}+x^n

\to (1-x)^n=1-^nC_1x+^nC_2x^2+...+(-1)^r ^nC_rx^r+...+(-1)^{n-1} ^nC_{n-1}x^{n-1}+(-1)^nx^n

General term:

If t_1, t_2,...,t_n,t_{n+1} are the terms in the expansion of (a+b)^n then the (r+1)^{th} term is called the General term.


Equidistant Term:

It can be verified that (r+1)^{th} term form the beginning =t_{r+1}=^nC_ra^{n-r}b^r and the (r+1)^{th} term form the end which  is equal to the (n+1-r)^{th} term from the beginning


Since ^nC_r=^nC_{n-r} , the coefficients of the equidistant terms from the beginning and end are equal.

Middle Term:

In the binomial expansion of  (x+y)^n there are (n+1) terms. Thus, there is only one middle term  n is even and there are two middle term if  n is odd.

(i) Let  n be even say n=2m  Then the middle term is equal to

 t_{m+1}=^{2m}C_m x^my^m

i.e.,  t_{\frac{n}{2}+1}=^nC_{\frac{n}{2}}x^{\frac{n}{2}}y^{\frac{n}{2}}

(ii) Let  n be odd say n=2m+1 , then

\to t_{m+1}=^{2m+1}C_m x^{m+1}y^m

i.e.,t_{\frac{n+1}{2}}=^nC_{\frac{n-1}{2}} x^{\frac{n+1}{2}}y ^{\frac{n-1}{2}}

\to t_{m+2}=^{2m+1}C_{m+1}x^m y^{m+1}

i.e.,t_{\frac{n+3}{2}}=^nC_{\frac{n+1}{2}}x^{\frac{n-1}{2}} y^{\frac{n+1}{2}}

 \therefore ^nC_{\frac{n-1}{2}}


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