Mathematical Induction problem. ?

Prove that #n cdot 1+(n-1) cdot2 +(n-2) cdot 3+...+2 cdot (n-1)+1 cdot n=1/6n(n+1)(n+2)#

2 Answers
Mar 12, 2018

See below.

Explanation:

#S_n = sum_(k=0)^n (n-k)(k+1)#
#S_n = 1/6n(n+1)(n+2)#

For #n=1#

#S_1 = 1#
#1/6 1 xx 2 xx 3 = 1#

Now assuming it is true for #n# we have for #n+1#

#S_(n+1) = sum_(k=0)^(n+1)(n+1-k)(k+1) = #

#= sum_(k=0)^n (n-k)(k+1)+sum_(k=0)^(n+1)(k+1) = #

#= 1/6n(n+1)(n+2)+((n+1)(n+2))/2 =#

#=1/6(n+1)(n+2)(n+3)#

so the statement is true.

Mar 13, 2018

Please go through The Explanation.

Explanation:

Let us prove the Result without using the Induction:

#"The Reqd. Sum="sum_(m=1)^(m=n)(n-m+1)m#,

#=sum{(n+1)m-m^2}#,

#=(n+1)sum_(m=1)^(m=n)m-sum_(m=1)^(m=n)m^2#,

#=(n+1){n/2(n+1)}-n/6(n+1)(2n+1)#,

#=n/6(n+1){3(n+1)-(2n+1)}#,

#=n/6(1+1)(n+2)#.