What is the coefficient #a_6# in #(1+x)^21+cdots+(1+x)^30# ?

3 Answers
Aug 13, 2017

Answer:

#""_21C_6+""_22C_6+...+""_30C_6.#

Explanation:

Recall that, #(1+x)^n=sum_(r=0)^(r=n)""_nC_rx^r.#

Therefore, the co-eff. of #x^r" in the expansion of "(1+x)^n# is #""_nC_r.#

Hence, the co-effs. of #x^6# in #(1+x)^n, where, n=21,22,...,30,#

are, #""_21C_6, ""_22C_6, ..., ""_30C_6,# resp.

#:.#The Reqd. Co-eff.=#""_21C_6+""_22C_6+...+""_30C_6.#

Aug 13, 2017

Answer:

See below.

Explanation:

#(1+x)^21+cdots+(1+x)^30=(1+x)^21(1+(1+x)+cdots+(1+x)^9)=#

#=(1+x)^21((1+x)^10-1)/((1+x)-1) = 1/x((1+x)^31-(1+x)^21)#

Now

#(1+x)^n=sum_(k=1)^n ((k),(n))x^k# then

#a_6 = ((7),(31))-((7),(21)) = (31!)/(7! (31 - 7)!) - (21!)/(7! (21 - 7)!) = 2513295#

Aug 13, 2017

Answer:

# Coef(x^6) = 2513295#

Explanation:

We seek the coefficient of #x^6# in:

# S = (1+x)^21 + (1+x)^22 + ... + (1+x)^30 #

Consider the sum

# S_n = (1+x)^0 + (1+x)^1 + (1+x)^2 + ... + (1+x)^n #

These term for a GP with #(n+1)# terms:

# a = 1#
# r = 1+x #

And so we can use the GP sum formula to evaluate the sum of the first #n# terms:

# S = a(1-r^n)/(1-r) #
# \ \ = (1-(1+x)^n)/(1-(1+x)) #
# \ \ = ((1+x)^n-1)/x #

Then we can find #S# using:

# S = S_31 - S_21 #

# \ \ = ((1+x)^31-1)/x - ((1+x)^21-1)/x #

# \ \ = (1+x)^31/x- (1+x)^21/x #

We can determine the coefficient of #x^r# in a binomial expansion #(1+x)^n# using the combination:

# ""_nC^r = ( (n), (r) ) = (n!)/((n-r)!r!)#

We can evaluate the combination directly using factorials, or more likely using a calculators in-built function for combinations.

As dividing by #x# will reduce the power of #x# in the binomial expansion then the coefficient of #x^6# that we seek is given by

# Coef(x^6) = Coef(x^7){(1+x)^31} - Coef(x^7){(1+x)^21} #
# " " = ( (31), (7) ) - ( (21), (7) )#
# " " = 26295755 - 116280#
# " " = 2513295#