Calculate #sum_(k=1)^(n-1) k(n-k+1) =# ?

2 Answers
Cesareo R.
Jan 26, 2018

See below.

Explanation:

You can use finite induction. So

#sum_(k=1)^(n-1) k(n-k+1) = (n^3+3n^2-4n)/6#

is true for #n = 2# when we have

#2 = 2#

Now supposing it is true for #n = m# we must proof that it remains true for #n = m+1# so

#sum_(k=1)^(m-1) k(m-k+1) = (m^3+3m^2-4m)/6#

and

#sum_(k=1)^(m) k(m-k+2) = sum_(k=1)^(m-1) k(m-k+1)+((m+1)(m+2))/2-1#

but

#(m^3 + 3 m^2 - 4 m)/6+((m+1)(m+2))/2-1 = ((m + 1)^3 + 3 (m + 1)^2 - 4 (m + 1))/6#

so the assertion is true.

Ratnaker Mehta
Jan 26, 2018

Kindly refer to the Explanation.

Explanation:

Suppose that,

#S_n=1(n)+2(n-1)+3(n-2)+...+(n-1)(2)+n(1)#.

Then, the reqd. sum #s,# is given by, #s=S_n-n(1).......(square)#.

If we write #S_n=sum_(m=1)^(m=n)T_m#, then observe that,

#T_1=1(n)=1(n-0)=1(n-(1-1))#,

#T_2=2(n-1)=2(n-(2-1))#,

#T_3=3(n-2)=3(n-(3-1))#,

#vdots vdots vdots vdots vdots vdots vdots vdots vdots vdots #

# rArr T_m=m(n-(m-1)=n(n-m+1)#.

#:. S_n=sum_(m=1)^(m=n)m(n-m+1)#,

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

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

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

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

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

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

#:. (square) rArr" The reqd. sum "s=n/6(n+1)(n+2)-n#,

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

#=n/6(n^2+3n+2-6)#,

#=(n^3+3n^2-4n)/6#.

Enjoy Maths., and Spread the Joy!

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