Question

Let $f(n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n}$ such that $P(n)f(n+2) = P(n)f(n) + q(n)$. Where $P(n)$, $Q(n)$ are polynomials of least possible degree and $P(n)$ has leading coefficient unity. Then match the following Column-I with Column-II.

• A. $\sum_{n=1}^{m} \frac{p(n)-2}{n}$
• B. $\sum_{n=1}^{m} \frac{q(n)-3}{2}$
• C. $\sum_{n=1}^{m} \frac{p(n) + q^2(n)-11}{n}$
• D. $\sum_{n=1}^{m} \frac{q^2(n) - p(n)-7}{n}$

Answer 1

A → D, B → A, C → B, D → C.

Answer 2

Solution Explanation

1. We are given
     $f(n)=1+\tfrac{1}{2}+\cdots+\tfrac{1}{n}$
     and
     $P(n)f(n+2)=P(n)f(n)+q(n).$
   Rearrange to get
     $q(n)=P(n)\Bigl(f(n+2)-f(n)\Bigr)=P(n)\left(\frac{1}{n+1}+\frac{1}{n+2}\right).$

2. To have $q(n)$ as a polynomial (and $P(n)$ of least degree with leading coefficient 1), choose
     $$ P(n)=(n+1)(n+2)=n^2+3n+2.

Then   $$ q(n)=(n+1)(n+2)\left(\frac{1}{n+1}+\frac{1}{n+2}\right)=(n+2)+(n+1)=2n+3.

3. Now compute the four sums (using $p(n)=P(n)=n^2+3n+2$ and $q(n)=2n+3$):

   • Row A:

     $$\sum_{n=1}^{m}\frac{p(n)-2}{n}=\sum_{n=1}^{m}\frac{(n^2+3n+2)-2}{n}=\sum_{n=1}^{m}\frac{n^2+3n}{n}=\sum_{n=1}^{m}(n+3).$$

     Thus,

     $$\sum_{n=1}^{m}(n+3)=\frac{m(m+1)}{2}+3m=\frac{m(m+1+6)}{2}=\frac{m(m+7)}{2}.$$

   • Row B:

     $$\sum_{n=1}^{m}\frac{q(n)-3}{2}=\sum_{n=1}^{m}\frac{(2n+3)-3}{2}=\sum_{n=1}^{m}\frac{2n}{2}=\sum_{n=1}^{m} n=\frac{m(m+1)}{2}.$$

   • Row C:
     First note $q^2(n)=(2n+3)^2=4n^2+12n+9$. Then,

     $$p(n)+q^2(n)-11=(n^2+3n+2)+(4n^2+12n+9)-11=5n^2+15n.$$

     Dividing by $n$,

     $$\frac{5n^2+15n}{n}=5n+15.$$

     So,

     $$\sum_{n=1}^{m}(5n+15)=5\sum_{n=1}^{m}(n+3)=5\left(\frac{m(m+1)}{2}+3m\right)=\frac{5m(m+7)}{2}.$$

   • Row D:
     We have,

     $$q^2(n)-p(n)-7=(4n^2+12n+9)-(n^2+3n+2)-7=3n^2+9n.$$

     Dividing by $n$,

     $$\frac{3n^2+9n}{n}=3n+9.$$

     Thus,

     $$\sum_{n=1}^{m}(3n+9)=3\sum_{n=1}^{m}n+9m=\frac{3m(m+1)}{2}+9m=\frac{3m(m+7)}{2}.$$

4. Now compare each computed sum with the given Column-II expressions:

   • Row A: Computed $\frac{m(m+7)}{2}$ matches Column-II option $\frac{m(m+7)}{2}$ (Row D option in the given table).
   • Row B: Computed $\frac{m(m+1)}{2}$ matches Column-II option $\frac{m(m+1)}{2}$ (Row A option).
   • Row C: Computed $\frac{5m(m+7)}{2}$ matches Column-II option $\frac{5m(m+7)}{2}$ (Row B option).
   • Row D: Computed $\frac{3m(m+7)}{2}$ matches Column-II option $\frac{3m(m+7)}{2}$ (Row C option).

Thus, the matching is:
  A → D, B → A, C → B, D → C.

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Final Answer:
A → D, B → A, C → B, D → C.