Question 941377
the sum of n terms of 1²/1 + (1²+2²)/2 + (1²+2²+3²)/3+.......is
a)(n/36)(4n²+15n+17) b) (n/36)(4n²-15n-17) c)(n/36)(4n²-15n+17) d) none
<pre>
{{{sum((expr(1/k)*sum((i^2),1=1,k)),k=1,n)}}}

Substituting sum of first k squares: {{{sum((i^2),1=1,k)=k(k+1)(2k+1)/6}}}



{{{sum(expr(1/k)*

(k(k+1)(2k+1)/6)


,k=1,n)}}}


{{{sum(expr(1/cross(k))*

(cross(k)(k+1)(2k+1)/6)


,k=1,n)}}}

{{{sum(

((k+1)(2k+1)/6)


,k=1,n)}}}


{{{expr(1/6)sum(

((k+1)^""(2k+1))


,k=1,n)}}}


{{{expr(1/6)sum(

(2k^2+3k+1)


,k=1,n)}}}

{{{expr(1/6)(sum(

(2k^2)


,k=1,n)+sum(

(3k)


,k=1,n)+sum(

(1)


,k=1,n)))}}}

Since {{{sum((1),k=1,n)=n}}}, and factoring out constants,

{{{expr(1/6)(2*sum(

(k^2)


,k=1,n)+3*sum(

(k)


,k=1,n)+n))}}}

Since {{{sum((k^2),k=1,n)=n(n+1)(2n+1)/6}}}, and {{{sum((k),1=1,n)=n(n+1)/2}}}


{{{expr(1/6)(2*expr(n(n+1)(2n+1)/6)




+3*expr(n(n+1)/2)+n))}}}


{{{expr(1/6)(2n(n+1)(2n+1)/6




+3n(n+1)/2+n))}}}

Getting an LCD

{{{expr(1/6)(2n(n+1)(2n+1)/6




+9n(n+1)/6+6n/6))}}}

{{{expr(1/6)( (2n(n+1)(2n+1)+9n(n+1)+6n)/6))}}}

Factor out n in the top, multiplying the two 6 denominators:

{{{ (n(2^""(n+1)(2n+1)+9(n+1)+6))/36)}}}

{{{ expr(n/36)(2^""(n+1)(2n+1)+9(n+1)+6))}}}

{{{ expr(n/36)(2^""(2n^2+3n+1)+9n+9+6))}}}

{{{ expr(n/36)(4n^2+6n+2+9n+15)}}}

{{{ expr(n/36)(4n^2+15n+17)}}}

Edwin</pre>