Please Help

[Guest]

Please Solve this, I am unable. Please, I beg

The question is attached,

I need d - f

Please Help

[shyamjayakannan]

d) Since [tex]n[/tex] is a multiple of [tex]m[/tex], let [tex]n=km[/tex]. Now, we can divide [tex]S_n^{(m)}[/tex] into [tex]m[/tex] different sums:
[tex]s_1=1+(m+1)+(2m+1)+...+\{(k-1)m+1\}[/tex]
[tex]s_2=2[2+(m+2)+(2m+2)+...+\{(k-1)m+2\}][/tex]
[tex]s_3=3[3+(m+3)+(2m+3)+...+\{(k-1)m+3\}][/tex]
and so on until
[tex]s_m=m[m+(m+m)+(2m+m)+...+\{(k-1)m+m\}][/tex]

Now, each of the smaller sums is an A.P with a common difference of [tex]im[/tex], first term [tex]i^2[/tex], and [tex]k[/tex] terms, [tex]i={1,2,3...,m}[/tex]. So we can use the formula for the sum of A.P: [tex]s=\frac{n}{2}\{2a+d(n-1)\}[/tex]

Using this, we get:
[tex]s_1=\frac{k}{2}\{2+m(k-1)\}[/tex], [tex]s_2=2\frac{k}{2}\{4+m(k-1)\}[/tex] and so on until [tex]s_m=m\frac{k}{2}\{2m+m(k-1)\}[/tex].

So, the final sum becomes:
[tex]S_n^{(m)}=\sum_{i=1}^mi\frac{k}{2}\{2i+m(k-1)\}=\sum_{i=1}^m\left\{i^2k+\frac{imk(k-1)}{2}\right\}=\frac{km(m+1)(2m+1)}{6}+\frac{km^2(m+1)(k-1)}{4}[/tex]

Now, replace [tex]k[/tex] with [tex]\frac{n}{m}[/tex] to get [tex]S_n^{(m)}=\frac{n(m+1)(2m+1)}{6}+\frac{n(m+1)(n-m)}{4}=\frac{n(m+1)}{2}\left(\frac{2m+1}{3}+\frac{n-m}{2}\right)=\boxed{\frac{n(m+1)(3n+m+2)}{12}}[/tex].

e) We can rewrite [tex]S_n^{(m)}[/tex] as follows:

[tex]S_n^{(m)}=\frac{n(m+1)(3n+m+2)}{12}=\frac{n(m+1)(3n+3m+2-2m)}{12}=\frac{3n(m+1)(n+m)-2n(m+1)(m-1)}{12}[/tex]

[tex]\displaystyle=\frac{(m+1)n(n+m)}{4}-\frac{n\left(m^2-1\right)}{6}=\frac{m^2(m+1)}{2}\times\frac{\displaystyle\frac{n}{m}\left(\frac{n}{m}+1\right)}{2}-\frac{n\left(m^2-1\right)}{6}=I_mS_{n/m}-nY_m[/tex].

From this, we get: [tex]\boxed{I_m=\frac{m^2(m+1)}{2}}[/tex] and [tex]\boxed{Y_m=\frac{m^2-1}{6}}[/tex]