Answer The Following Question With Explanation ?

#(1^2)+(1^2+2^2)+(1^2+2^2+3^2)+............#
Summation Of #n^{th)# term

2 Answers
Ratnaker Mehta
Aug 1, 2018

# n/12(n+1)^2(n+2)#.

Explanation:

Observe that, the #m^(th)# term #t_m# is given by,

#t_m=1^2+2^2+...+m^2=m/6(m+1)(2m+1)#.

#:."The Reqd. Sum "s_n=sum_(m=1)^(m=n)t_m#,

#=sum{m/6(m+1)(2m+1)}#,

#=1/6sum{2m^3+3m^2+m}#,

#=1/6{2summ^3+3summ^2+summ}#,

#=2/6*n^2/4(n+1)^2+3/6*n/6(n+1)(2n+1)+1/6*n/2(n+1)#,

#=n^2/12(n+1)^2+n/12(n+1)(2n+1)+n/12(n+1)#,

#=n/12(n+1)[{n^2+n}+{2n+1}+1]#,

#=n/12(n+1){(n^2+3n+2)}#,

#=n/12(n+1){(n+1)(n+2)}#.

#rArr"The Reqd. Sum"=n/12(n+1)^2(n+2)#.

maganbhai P.
Aug 1, 2018

#S_n=n/12(n+1)^2(n+2)#

Explanation:

Let ,

#S#=#1^2+(1^2+2^2)+(1^2+2^2+3^2)+...+(1^2+2^2+...+n^2)#

#:.t_n=1^2+2^2+3^2+...+n^2#

#:.t_n=sum_(r=1)^n r^2=n/6(n+1)(2n+1)#

#:.t_n=1/6 {n(2n^2+2n+n+1)}=1/6{2n^3+2n^2+n^2+n}#

#:.t_n=1/6{2n^3+3n^2+n}#

#:.S_n=1/6{2sum_(r=1)^nr^3+3sum_(r=1)^n r^2+sum_(r=1)^n r}#

#=1/6{cancel2n^2/cancel4^2(n+1)^2+cancel3n/cancel6^2(n+1)(2n+1)+n/2(n+1)}#

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

#=n/12(n+1){n^2+n+2n+2}#

#=n/12(n+1)(n^2+3n+2)#

#:.S_n=n/12(n+1)(n+1)(n+2)#

#:.S_n=n/12(n+1)^2(n+2)#
........................................................................................
Note:
#(1)sum_(r=1)^n r=n/2(n+1)#
#(2)sum_(r=1)^n r^2=n/6(n+1)(2n+1)#
#(3)sum_(r=1)^nr^3=n^2/4(n+1)^2#

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