Question

Let $S_n = \sum_{r=2}^{n} \frac{3^{r-1}(2r-3)}{r(r-1)}$, then

• A. $S_9$ is divisible by 4
• B. $S_9$ is divisible by 21
• C. $10.S_{10}+3 = 3^{10}$
• D. $7(S_7+3), 10(S_{10}+3), 13(S_{13}+3)$ are in G.P.

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

To evaluate the sum $S_n = \sum_{r=2}^{n} \frac{3^{r-1}(2r-3)}{r(r-1)}$, we first analyze the general term $T_r = \frac{3^{r-1}(2r-3)}{r(r-1)}$.

Step 1: Decompose the rational part of $T_r$ using partial fractions.

Let $\frac{2r-3}{r(r-1)} = \frac{A}{r-1} + \frac{B}{r}$. Multiplying by $r(r-1)$, we get $2r-3 = A r + B (r-1)$. To find $A$, set $r=1$: $2(1)-3 = A(1) + B(0) \implies -1 = A$. To find $B$, set $r=0$: $2(0)-3 = A(0) + B(-1) \implies -3 = -B \implies B=3$. So, $\frac{2r-3}{r(r-1)} = \frac{-1}{r-1} + \frac{3}{r}$.

Step 2: Rewrite $T_r$ in a telescoping form.

Substitute the partial fraction decomposition back into $T_r$: $T_r = 3^{r-1} \left( \frac{3}{r} - \frac{1}{r-1} \right)$ $T_r = \frac{3 \cdot 3^{r-1}}{r} - \frac{3^{r-1}}{r-1}$ $T_r = \frac{3^r}{r} - \frac{3^{r-1}}{r-1}$

Let $f(r) = \frac{3^r}{r}$. Then $T_r = f(r) - f(r-1)$. This is a telescoping form.

Step 3: Calculate the sum $S_n$.

$S_n = \sum_{r=2}^{n} T_r = \sum_{r=2}^{n} \left( f(r) - f(r-1) \right)$ $S_n = (f(2) - f(1)) + (f(3) - f(2)) + \dots + (f(n) - f(n-1))$ All intermediate terms cancel out, leaving: $S_n = f(n) - f(1)$ $S_n = \frac{3^n}{n} - \frac{3^1}{1}$ $S_n = \frac{3^n}{n} - 3$.

Step 4: Check the given options.

(A) $S_9$ is divisible by 4. $S_9 = \frac{3^9}{9} - 3 = \frac{3^9}{3^2} - 3 = 3^7 - 3$. $S_9 = 3(3^6 - 1)$. We know $3^6 = (3^3)^2 = 27^2 = 729$. $S_9 = 3(729 - 1) = 3(728)$. To check divisibility by 4, we check if 728 is divisible by 4. $728 = 4 \times 182$. So, $S_9 = 3 \times (4 \times 182) = 12 \times 182$. Since $S_9$ contains a factor of 4, it is divisible by 4. Option (A) is correct.

(B) $S_9$ is divisible by 21. From (A), $S_9 = 3(728)$. For $S_9$ to be divisible by 21, it must be divisible by 3 and 7. It is clearly divisible by 3. Now check divisibility by 7: $728 \div 7 = 104$. So, $S_9 = 3 \times (7 \times 104) = 21 \times 104$. Since $S_9$ contains a factor of 21, it is divisible by 21. Option (B) is correct.

(C) $10 \cdot S_{10} + 3 = 3^{10}$. Using the formula $S_n = \frac{3^n}{n} - 3$: $S_{10} = \frac{3^{10}}{10} - 3$. $10 \cdot S_{10} = 10 \left( \frac{3^{10}}{10} - 3 \right) = 3^{10} - 30$. Then $10 \cdot S_{10} + 3 = (3^{10} - 30) + 3 = 3^{10} - 27$. Since $3^{10} - 27 \neq 3^{10}$, option (C) is incorrect.

(D) $7(S_7+3), 10(S_{10}+3), 13(S_{13}+3)$ are in G.P. From $S_n = \frac{3^n}{n} - 3$, we can write $S_n + 3 = \frac{3^n}{n}$. Multiplying by $n$, we get $n(S_n+3) = 3^n$. Let's find the terms: For $n=7$: $7(S_7+3) = 3^7$. For $n=10$: $10(S_{10}+3) = 3^{10}$. For $n=13$: $13(S_{13}+3) = 3^{13}$. The sequence of terms is $3^7, 3^{10}, 3^{13}$. To check if they are in G.P., we find the common ratio: Ratio 1: $\frac{3^{10}}{3^7} = 3^{10-7} = 3^3 = 27$. Ratio 2: $\frac{3^{13}}{3^{10}} = 3^{13-10} = 3^3 = 27$. Since the common ratio is constant (27), the terms are in G.P. Option (D) is correct.