`text(Propertices of)` `text()^nC_r`
`(1)` `text()^nC_r=text()^nC_(n-r)=>text()^nC_x=text()^nC_y` Cy has two solution `x = y` or `x + y = n`.
`(2)` `text()^nC_r+text()^nC_(r-1)=text()^(n-1)C_r`
`(3)` `text()^nC_r=n/r (text()^(n-1)C_(r-1))`
`(4)` `(text()^nC_r)/(text()^nC_(r-1))=(n-r+1)/r`
ln the expansion of `(1+x)^n` ; i.e., `(1+x)^n=text()^nC_0+text()^nC_1x+...............+ text()^nC_rx^r+.............+text()^nC_nx^n`
The coefficients `text()^nC_0 , text()^nC_1 , text()^nC_n` of various powers of `x`, are called binomial coefficients and they are written as
`C_0,C_1,C_2,.............C_n`
Hence
`(1+x)^n=C_0+C_1x+C_2x^2+...............+C_rx^r+................+C_nx^n`.........................`(1)`
Where `C_0=1,C_1=n,C_2=(n(n-1))/(2!)`
`C_r=(n(n-1)...........(n-r+1))/(r!) , C_n=1`
Now, we shall obtain some important expressions involving binomial coefficients-
`(a) text( Sum of Coefficient: putting x = 1 in (1))`, we get
`C_0+C_1+C_2+.........+ C_n=2^n`.....................`(2)`
`(b) text( Sum of coefficients with alternate signs: putting x =- 1 in (1))` we get
`C_0-C_1+C_2-C_3+............=0`................(3)
`(c) text( Sum of coefficients of even and odd terms: from (3))`, we have
`C_0+C_2+C_4+.................=C_1+C_3+C_5+......................`.............(4)
i.e. sum of coefficients of even and odd terms are equal.
from `(2)` and `(4)`
`=> C_0+C_2+.............=C_1+C_3+.............=2^(n-1)`
`(d) text( Sum of products of coefficients: Replacing x by)` `1//x` in `( 1)`
We get
`(1+1/x)^n=C_0+C_1/x+C_2/x+............+C_n/x^n+...................`...............(5)
Multiplying `(1)` by `(5)`, we get
`((1+x)^(2n))/x^n =(C_0+C_1x+C_2x^2+.........)(C_0+C_1/x+C_2/x^2+................)`
Now, comparing coefficients of `x^r` on both the sides, we get
`= (2n !)/((n+r)!(n-r)!)`................(6)
`(e) text( Sum of squares of coefficients :)`
putting `r= 0` in `(6)`, we get
`C_0^2+C_1^2+C_2^2+...................+C_n^2=(2n!)/(n!n!)`
`(f) text( putting r= 1 in (6))`, we get
`C_0C_1+C_1C_2+C_2C_3+.............+C_(n-1)C_n=text()^(2n)C_(n-1)`
`=(2n!)/((n+1)!(n-1)!)`..............(7)
`(g) text( putting r= 2 in (6))` , we get
`C_0C_2 + C_1C_3 + C_2C_4 + .... + C_(n-2)C_n=text()^(2n)C_(n-2)`
`=(2n!)/((n+2)!(n-2)!)`.....................(8)
`(h) text( Use of Differentiation :)`
`(1+x)^n=C_0C_1x+C_2x^2+.................+C_rx^r+.............+C_nx^n`.................(1)
Differentiating both sides of `( 1)` w.r.t. `x`, we get
`n(1+x)^(n-1)=C_1+2C_2x+3C_3x^2+..............+nC_nx^(n-1)`
Now putting `x = 1 ` and `x =- 1` respectively
`C_1+ 2C_2 +3C_3+ ..... + nC_n = n . 2^(n-1)`................(9)
and `C_1 - 2C_2+ 3C_3 - ....... = 0` ..................(10)
`(i) text( Adding) ``(2)` `text( and )``(9)`
`C_0+2C_1+3C_2+..............+(n+1)C_n=2^(n-1)(n+2)`.....................(11)
`(j) text( Use of Integration :)`
`( 1+x)^n = C_0+C_1x + C_2x^2+ .... + C_rx^r + ...... + C_nx^n`....................(1)
Integrating `( 1)` w. r.t. `x` between the limits `0` to `1` , we get,
`[((1+x)^(n+1))/(n+1)]_0^1=[C_0x+C_1x^2/2+C_2/2+............+(C_nx^(n+1))/(n+1)]`
`=> C_0+C_1/2+C_2/3+......+C_n/(n+1)=(2^(n+1)-1)/(n+1)`................(12)
Integrating `(1)` w.r.t. `x` between the limits `- 1` to `0`, we get
`[((1+x)^(n+1))/(n+1)]_(-1)^0=[C_0x+C_1x^2/2+C_2/2+............+(C_nx^(n+1))/(n+1)]_(-1)^0`
`=> C_0-C_1/2+C_2/3-C_3/4+................+((-1)^n . C_n)/(n+1)=1/((n+1))`.............(13)
`text(Point to consider : )`
1. In the expansion of `(x- 2y + 3z)^n`, putting` x = y = z = 1,` then we get the sum of coefficients `= (1 -2 + 3)^n = 2^n.`
2. In the expansion of `(1 + x + x^2)^n`, putting x = 1, then we get the sum coefficients `= (1 + 1 + 1)^n = 3^n.`
`text(Propertices of)` `text()^nC_r`
`(1)` `text()^nC_r=text()^nC_(n-r)=>text()^nC_x=text()^nC_y` Cy has two solution `x = y` or `x + y = n`.
`(2)` `text()^nC_r+text()^nC_(r-1)=text()^(n-1)C_r`
`(3)` `text()^nC_r=n/r (text()^(n-1)C_(r-1))`
`(4)` `(text()^nC_r)/(text()^nC_(r-1))=(n-r+1)/r`
ln the expansion of `(1+x)^n` ; i.e., `(1+x)^n=text()^nC_0+text()^nC_1x+...............+ text()^nC_rx^r+.............+text()^nC_nx^n`
The coefficients `text()^nC_0 , text()^nC_1 , text()^nC_n` of various powers of `x`, are called binomial coefficients and they are written as
`C_0,C_1,C_2,.............C_n`
Hence
`(1+x)^n=C_0+C_1x+C_2x^2+...............+C_rx^r+................+C_nx^n`.........................`(1)`
Where `C_0=1,C_1=n,C_2=(n(n-1))/(2!)`
`C_r=(n(n-1)...........(n-r+1))/(r!) , C_n=1`
Now, we shall obtain some important expressions involving binomial coefficients-
`(a) text( Sum of Coefficient: putting x = 1 in (1))`, we get
`C_0+C_1+C_2+.........+ C_n=2^n`.....................`(2)`
`(b) text( Sum of coefficients with alternate signs: putting x =- 1 in (1))` we get
`C_0-C_1+C_2-C_3+............=0`................(3)
`(c) text( Sum of coefficients of even and odd terms: from (3))`, we have
`C_0+C_2+C_4+.................=C_1+C_3+C_5+......................`.............(4)
i.e. sum of coefficients of even and odd terms are equal.
from `(2)` and `(4)`
`=> C_0+C_2+.............=C_1+C_3+.............=2^(n-1)`
`(d) text( Sum of products of coefficients: Replacing x by)` `1//x` in `( 1)`
We get
`(1+1/x)^n=C_0+C_1/x+C_2/x+............+C_n/x^n+...................`...............(5)
Multiplying `(1)` by `(5)`, we get
`((1+x)^(2n))/x^n =(C_0+C_1x+C_2x^2+.........)(C_0+C_1/x+C_2/x^2+................)`
Now, comparing coefficients of `x^r` on both the sides, we get
`= (2n !)/((n+r)!(n-r)!)`................(6)
`(e) text( Sum of squares of coefficients :)`
putting `r= 0` in `(6)`, we get
`C_0^2+C_1^2+C_2^2+...................+C_n^2=(2n!)/(n!n!)`
`(f) text( putting r= 1 in (6))`, we get
`C_0C_1+C_1C_2+C_2C_3+.............+C_(n-1)C_n=text()^(2n)C_(n-1)`
`=(2n!)/((n+1)!(n-1)!)`..............(7)
`(g) text( putting r= 2 in (6))` , we get
`C_0C_2 + C_1C_3 + C_2C_4 + .... + C_(n-2)C_n=text()^(2n)C_(n-2)`
`=(2n!)/((n+2)!(n-2)!)`.....................(8)
`(h) text( Use of Differentiation :)`
`(1+x)^n=C_0C_1x+C_2x^2+.................+C_rx^r+.............+C_nx^n`.................(1)
Differentiating both sides of `( 1)` w.r.t. `x`, we get
`n(1+x)^(n-1)=C_1+2C_2x+3C_3x^2+..............+nC_nx^(n-1)`
Now putting `x = 1 ` and `x =- 1` respectively
`C_1+ 2C_2 +3C_3+ ..... + nC_n = n . 2^(n-1)`................(9)
and `C_1 - 2C_2+ 3C_3 - ....... = 0` ..................(10)
`(i) text( Adding) ``(2)` `text( and )``(9)`
`C_0+2C_1+3C_2+..............+(n+1)C_n=2^(n-1)(n+2)`.....................(11)
`(j) text( Use of Integration :)`
`( 1+x)^n = C_0+C_1x + C_2x^2+ .... + C_rx^r + ...... + C_nx^n`....................(1)
Integrating `( 1)` w. r.t. `x` between the limits `0` to `1` , we get,
`[((1+x)^(n+1))/(n+1)]_0^1=[C_0x+C_1x^2/2+C_2/2+............+(C_nx^(n+1))/(n+1)]`
`=> C_0+C_1/2+C_2/3+......+C_n/(n+1)=(2^(n+1)-1)/(n+1)`................(12)
Integrating `(1)` w.r.t. `x` between the limits `- 1` to `0`, we get
`[((1+x)^(n+1))/(n+1)]_(-1)^0=[C_0x+C_1x^2/2+C_2/2+............+(C_nx^(n+1))/(n+1)]_(-1)^0`
`=> C_0-C_1/2+C_2/3-C_3/4+................+((-1)^n . C_n)/(n+1)=1/((n+1))`.............(13)
`text(Point to consider : )`
1. In the expansion of `(x- 2y + 3z)^n`, putting` x = y = z = 1,` then we get the sum of coefficients `= (1 -2 + 3)^n = 2^n.`
2. In the expansion of `(1 + x + x^2)^n`, putting x = 1, then we get the sum coefficients `= (1 + 1 + 1)^n = 3^n.`