Binomial sikhna chahte ho to sikho vana koi baat nahin :P

Q1 Find the real values of a,b,p,q for which (2x-1)^{20}-(ax+b)^{20}=(x^2+px+q)^{20}

Q2 Find Sum of series S=C_1-(1+ 1/2)C_2 +(1+ 1/2 +1/3)C_3+...+(-1)^{n-1}(1+ 1/2+....+1/n)C_n

Q3 Sum of series S=\sum_{r=0}^{n}\frac{3^{r+4}.^nC_r}{^{r+4}C_r}+\sum_{r=0}^{3}{\frac{^{n+4}C_r.3^r}{^{n+4}C_4}}

Q4 Find S
S=1-C_1\frac{(1+x)}{(1+nx)}+C_2\frac{(1+2x)}{(1+nx)^2}-C_3\frac{(1+3x)}{(1+nx)^3}+....\ upto (n-1) terms
*it is (n+1) terms *
Q5 If n>2 then C_1(a-1)^2-C_2(a-2)^2+C_3(a-3)^2+..+(-1)^{n-1}.C_n(a-n)^2=????

5 Answers

24
eureka123 ·

Q6\sum_{k=1}^{n}{C_k.C_{k-1}}=??

Q7 Find value of S S=1.2.3.4+2.3.4.5+.....+n.(n+1).(n+2).(n+3)

1
xYz ·

@eureka is it n-1 terms or n+1 terms in Q4
\frac{d(x(1-x^{t})^{n})}{dx}=^{n}C_{0}-(1+t) ^{n} C_{1} +(1+2t) ^{n}C_{2}+............
*please include coefficients as xtr in front of nCr
=(1-x^{t})^{n-1}(1-x^{t}(1+nt))=S1
now consider g.p: S2 = 1+\frac{1}{((1+nx)x)^{r}}+\frac{1}{((1+nx)x)^{2r}}+\frac{1}{((1+nx)x)^{3r}}+.......................
answer is coefficient of x0 in the expansion of S1S2

24
eureka123 ·

edited

62
Lokesh Verma ·

Q 6) 2nCn+1

106
Asish Mahapatra ·

Q7. Tn = n(n+1)(n+2)(n+3) = [n+4-(n-1)]n(n+1)(n+2)(n+3)/5
= 15[(n)(n+1)(n+2)(n+3)(n+4) - (n-1)(n)(n+1)(n+2)(n+3)]

Summing we get S = 15[(n)(n+1)(n+2)(n+3)(n+4)

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