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
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).

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)
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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