JEE Mains · Maths · STD 11 - 8. sequence and series
Let the first term of a series be \(T_1=6\) and its \(\mathrm{r}^{\text {th }}\) term \(T_r=3 T_{r-1}+6^r, r=2,3, \ldots . ., n\). If the sum of the first \(\mathrm{n}\) terms of this series is \(\frac{1}{5}\left(n^2-12 n+39\right)\) \(\left(4.6^n-5.3^n+1\right)\). Then \(n\) is equal to ...........
- A \(10\)
- B \(5\)
- C \(6\)
- D \(11\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\( \mathrm{T}_{\mathrm{r}}=3 \mathrm{~T}_{\mathrm{r}-1}+6^{\mathrm{r}}, \mathrm{r}=2,3,4, \ldots \mathrm{n} \) \( \mathrm{T}_2=3 \cdot \mathrm{T}_1+6^2 \) \( \mathrm{~T}_2=3 \cdot 6+6^2 \) ................(\(1\)) \( \mathrm{~T}_3=3 \mathrm{~T}_2+6^3 \)…
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