If the three terms of the expansion of (ltPx)^n in ascending powers of x are 1+20x+160x^2, find the values if n and p?

1 Answer

The answer is #n=5# and #p=4#

Explanation:

#(1+px)^n=1+20x+160x^2+....#

By the binomial theorem,

#(1+px)^n=1+npx+(n(n-1))/(1*2)p^2x^2+.....#

Therefore,

#np=20#, #=>#, #p=20/n#

and

#(n(n-1)p^2)/2=160#

#(n(n-1)*400/n^2)/2=160#

#(n-1)400=320n#

#400n-400=320n#

#400n-320n=400#

#80n=400#

#n=5#

and

#p=20/5=4#