Using binomial theorem (without using the formula for `^n C_r` ) , prove that `^n C_4+^m C_2-^m C_1^n C_2=^m C_4-^(m+n)C_1^m C_3+^(m+n)C_2^m C_2-^(m+n)C_3^m C_1+^(m+n)C_4dot`
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`.^(m)C_(4)-.^(m+n)C_(1).^(m)C_(3)+.^(m+n)C_(2).^(m)C_(2)-.^(m+n)C_(3).^(m)C_(1)+.^(m+n)C_(4)`
`= .^(m+n)C_(0).^(m)C_(4)-.^(m+n)C_(1).^(m)C_(3)+.^(m+n)C_(2).^(m)C_(2)-.^(m+n)C_(3).^(m)C_(1)+.^(m+n)C_(4).^(m)C_(0)`
= Coefficient of `x^(4)` in `(1+x)^(m+n)(1-x)^(m)`
= Coefficient of `x^(4)` in `(1-x)^(m)(1+x)^(n)`
= Coefficient of `x^(4)` in `[1-.^(m)C_(1)x^(2)+.^(m)C_(2)x^(4)-"....."][1+.^(n)C_(1)x+.^(n)C_(2)x^(2)+"....."+.^(n)C_(n)x^(n)]`
`= .^(n)C_(4)-.^(m)C_(1) xx .^(n)C_(2)+.^(m)C_(2)`
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Let `C_1 and C_2`, be the graph of the functions `y= x^2 and y= 2x, 0<=x<= 1` respectively. Let `C_3`, be the graph of a function `y- (fx), 0<=x<=1, f
Correct Answer - `f(x) = x^(3) - x^(2)`
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Prove that `^100 C_2^(100)C_2+^(100)C_2^(100)C_4+^(100)C_4^(100)C_6++^(100)C_(98)^(100)C_(100)=1/2[^(200)C_(98)-^(100)C_(49)]dot`
To find `S = .^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"....."+.^(100)C_(98).^(100)C_(100)` Consider, `.^(100)C_(0).^(100)C_(2)+.^(100)C_(1).^(100)C_(3) + .^(100)C_(2).^(100)C_(4)+.^(100)C_(3)+.^(100)C_(5)+"...." + .^(100)C_(98).^(100)C_(100)` `= .^(100)C_(0).^(100)C_(98)+.^(100)C_(1).^(100)C_(97) + .^(100)C_(2).^(100)C_(96)+.^(100)C_(3).^(100)C_(95) + "....."+^(100)C_(98) .^(100)C_(0)` = Coefficients of `x^(98)` in `(1+x)^(100) (1+x)^(100)` `= .^(200)C_(98) " "(1)` Also,...
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If `(18 x^2+12 x+4)^n=a_0+a_(1x)+a2x2++a_(2n)x^(2n),` prove that `a_r=2^n3^r(^(2n)C_r+^n C_1^(2n-2)C_r+^n C_2^(2n-4)C_r+)` .
`(18x^(2)+12x+4)^(n) = 2^(n)[2+9x^(2)+6x]^(n)` Now, `a_(r )` is coefficient of `x^(r )` in `2^(n) [(3x+1)^(2)+1]^(n)`. Hence `a_(r) =` Coefficient of `x^(r )2^(n)[.^(n)C_(0)(3x+1)^(2n)+.^(n)C_(1)(3x+1)^(2n-2) + .^(n)C_(2)(3x+1)^(2n-4)+"…."+.^(n)C_(r )(3x+1)^(2n-2r)+"....."]` or `a_(r)=2^(n)[.^(n)C_(0)3^(r).^(2n)C_(r)+.^(n)C_(1)3^(r).^(2n-2)C_(r)+.^(n)C_(2)3^(r).^(2n-4)C_(r)+"...."]` `= 2^(n)3^(r)[.^(n)C_(0).^(2n)C_(r)+.^(n)C_(1).^(2n-2)C_(r)+.^(n)C_(2).^(2n-4)C_(r)+"...."]`
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Find the value of `1/(81^n)-(10)/(81^n)^(2n)C_1+(10^2)/(81^n)^(2n)C_2-(10^3)/(81^n)^(2n)C_3++(10^(2n))/(81^n)` .
Correct Answer - `1` We have `(1)/(81^(n)) - (10)/(81^(n)).^(2n)C_(1)+(10^(2))/(81^(n)).^(2n)C_(2)- (10^(3))/(81^(n)) .^(2n)C_(3)+"……."+(10^(2n))/(81^(n))` `= 1/(81^(n))[.^(2n)C_(0) - .^(2n)C_(1)10^(1) + .^(2n)C_(2)10^(2)-.^(2n)C_(3)10^(3)+"……"+.^(2n)C_(2n)10^(2n)]` `= (1)/(81^(n))[1-10]^(2n)` `= ((-9)^(2n))/(81^(n))= (81^(n))/(81^(n)) = 1`
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Find the coefficient of `x^n` in the polynomial `(x+^n C_0)(x+3^n C_1)xx(x+5^n C_2)[x+(2n+1)^n C_n]dot`
Correct Answer - `(n+1)2^(n)` Given polynomial is `n +1` degree polynomial. `(x+.^(n)C_(0))(x+3.^(n)C_(2))(x+5.^(n)C_(2))"....."[x+(2n+1).^(n)C_(n)]` `= x^(n+1)+a_(1)x^(n)+"...."+a_(n+1)` Then, `- (a_(1))/(a_(0))=-.^(n)C_(0)-3.^(n)C_(1)-5.^(n)C_(2)-"....."-(2n+1).^(n)C_(n)` `rArr a_(1)=.^(n)C_(0)+3.^(n)C_(1)+5.^(n)C_(2)-"......"+(2n+1).^(n)C_(n)` `= underset(r=0)overset(n)sum.^(n)C_(r)(2r+1)=2underset(r=0)overset(n)sumr^(n)C_(r)+underset(r=0)overset(n)sum.^(n)C_(r)` `= 2underset(r=0)overset(n)sumn..^(n-1)C_(r-1) + underset(r=0)overset(n)sum.^(n)C_(r)` `= 2n 2^(n+1) + 2^(n) =...
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Prove that `1/(n+1)=(^n C_1)/2-(2(^n C_2))/3+(3(^n C_3))/4-+(-1)^(n+1)(n(^n C_n))/(n+1)` .
`S = (.^(n)C_(1))/(2)-(2(.^(n)C_(3)))/(3)+(3(.^(n)C_(3)))/(4)"...."+(-1)^(n+1)(n(.^(n)C_(n)))/(n+1)` ` = underset(r=1)overset(n)sum (r.^(n)C_(r))/((r+1))(-1)^(r+1)` `= underset(r=1)overset(n)sum r(.^(n+1)C(r+1))/(n+1)(-1)^(r+1)` `= 1/(n+1) underset(r=1)overset(n)sum [(r+1).^(n+1)C_(r+1)(-1)^(r+1)-.^(n+1)C_(r+1)(-1)^(r+1)]` `= (1)/(n+1)underset(r=1)overset(n)sum[-(n+1).^(n)C_(r)(-1)^(r)-.^(n+1)C_(r+1)(-1)^(r+1)]` `= -[-.^(n)C_(2)-.^(n)C_(3)+"...."+(-1)^(n).^(n)C_(n)]` ` - 1/(n+1)[.^(n+1)C_(2) - .^(n+1)C_(3)+".....(-1)^(n+1).^(n+1)C_(n+1)]` `= - [(1-1)^(n)-1]-(1)/(n+1)[(1-1)^(n+1) - .^(n+1)C_(0)+.^(n+1)C_(1)]` ` =...
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`(n+2)^n C_0 2^(n+1)-(n+1)^n C_1 2^n^n C_2 2^(n-1)-` is equal to `4` b. `4n` c. `4(n+1)` d. `2(n+2)`
Correct Answer - C `t_(r+1)=(-1)^(r)(n-r+2).^(n)C_(r)2^(n-r+1)` `= (n+2)2^(n+1)(-1)^(r).^(n)C_(r)(1/2)^(r)-2^(n+1)r.^(n)C_(r)(1/2)^(r)` `= (n+2)2^(n+1).^(n)C_(r)(-1/2)^(r) + 2^(n)n.^(n-1)C_(r-1)(-1/2)^(r-1)` `:.` Sum `= (n+2)2^(n+1){.^(n)C_(0) - .^(n)C_(1) xx 1/2 .^(n)C_(2)xx(1/2)^(2)-"...."}` `+2^(n){.^(n-1)C_(0)-.^(n-1)C_(1)xx1/2+.^(n-1)C_(2)xx(1/2)^(2)+"...."}` `= (n+2)2^(n+1)(1-1/2)^(n)+n2^(n)(1-1/2)^(n-1)` `= 2(n+2)+2n` `= 4n+4`
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If for `z` as real or complex, `(1+z^2+z^4)^8=C_0+C1z2+C2z4++C_(16)z^(32)t h e n` `C_0-C_1+C_2-C_3++C_(16)=1` `C_0+C_3+C_6+C_9+C_(12)+C_(15)=3^7` `C_2
Correct Answer - A::B::D `(1+z^(2)+z^(4))^(8) = C_(0) + C_(1)z^(2) + X_(2)z^(4) + "….." + C_(16)z^(32) " "(1)` Putting `x = i`, where `i = sqrt(-1)`. `(1-1+1)^(8) = C_(0) - C_(1) +...
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If `(1+x)^n=C_0+C_1x+C2x2++C_n x^n , n in N ,t h e nC_0-C_1+C_2-+(-1)^(n-1)C_(m-1),` is equal to `(m<n)` `((n-1)(n-2)(n-m+1))/((m-1)!)(-1)^(m-1)` `^n-
Correct Answer - A::B::D `((n-1)(n-2)"……"(n-m+1))/((m-1)!)` = `=((n-1)(n-2)"...."(n-m+1)(n-m)"...."2.1)/((n-m)!(m-1)!)` `=.^(n-1)C_(m-1)` `=` Coefficient of `x^(m-1)` in `(1+x)^(n-1)` `=` Coefficient of `x^(m-1)` in `(1+x)^(n)(1+x)^(-1)` Now, `(1+x)^(n) = C_(0)+C_(1)x+C_(2)x^(2)+"...."+C_(m-1)x^(m-1)+"...."+C_(n)x^(n)" "(1)` `(1+x)^(-1) = 1-x+x^(2)-x^(3)+"...."+(-1)^(m-1)x^(m-1)+"......"" "(2)` Collecting the...
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If `(a_1)/(a_2) = (b_1)/(b_2) ne (c_1)/(c_2)`, then the equations `a_1x + b_1y + c_1 =0 and a_2x + b_2y + c_2=0` are _________. (consistent/inconsiste
Correct Answer - Inconsistent
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