Find the sum of the series. 243+324+432+....... to n terms

Correct Answer is : 36( 4n3n−1)3^{6}≤ft(\;{4ⁿ}{3ⁿ}-1)36(3n4n−1)

Correct Answer is : 36( 4n3n−1)3^{6}≤ft(\;{4ⁿ}{3ⁿ}-1)36(3n4n−1)

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Question : 41 of 86

Marks: +1, -0

243+324+432+.......

n

terms

Solution:

Concept:

The series is a geometric progression (GP) with first term

243

and common ratio

\frac{4}{3}

Explanation:

First term

a = 243 = 3^5

Common ratio

r = \frac{324}{243} = \frac{4}{3}

Sum of

n

terms of GP:

S_n = a \frac{r^n - 1}{r - 1}

Substitute values:

S_n = 3^5 \cdot \frac{(\frac{4}{3})^n - 1}{\frac{4}{3} - 1} = 3^5 \cdot \frac{\frac{4^n}{3^n} - 1}{\frac{1}{3}} = 3^5 \cdot 3 \left( \frac{4^n}{3^n} - 1 \right) = 3^6 \left( \frac{4^n}{3^n} - 1 \right)

Answer:

3^6 \left( \frac{4^n}{3^n} - 1 \right)

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