Prove that- 3-2 2 - 3 2 - 2 3 - 3 3 - 2 4 - 3 n - 2 n-1 - 12 5 6 n -1 -

Given series forms a G.P with a=12 and r=6 S= a( r ^ n 1 / r 1 #160; ) #160; S=12( 6 ^ n 1 / 6 1 #160; ) S= 12 / 5 ( 6 ...

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Standard XII

Mathematics

Sum of Infinite Terms of a GP

Prove that: ...

Question

Prove that:
${3.2}^{2} + 3^{2} \cdot 2^{3} + 3^{3} \cdot 2^{4} + . . . + 3^{n} \cdot 2^{n + 1} = \frac{12}{5} (6^{n} - 1)$ .

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Solution

Given series forms a G.P with $a = 12$ and $r = 6$
$S = a (\frac{r^{n} - 1}{r - 1}) S = 12 (\frac{6^{n} - 1}{6 - 1}) S = \frac{12}{5} (6^{n} - 1)$

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Prove that:3.22+32⋅23+33⋅24+...+3n⋅2n+1=125(6n−1).

Given series forms a G.P with a=12 and r=6S=a(rn−1r−1)S=12(6n−16−1)S=125(6n−1)

Prove that:
${3.2}^{2} + 3^{2} \cdot 2^{3} + 3^{3} \cdot 2^{4} + . . . + 3^{n} \cdot 2^{n + 1} = \frac{12}{5} (6^{n} - 1)$ .

Given series forms a G.P with $a = 12$ and $r = 6$
$S = a (\frac{r^{n} - 1}{r - 1}) S = 12 (\frac{6^{n} - 1}{6 - 1}) S = \frac{12}{5} (6^{n} - 1)$