Subjective Type

Evaluate $$\displaystyle \sum _{ k=1 }^{ 11 }{ { (2+3 }^{ k }) } $$

Solution

$$\displaystyle \sum _{ k=1 }^{ 11 }{ (2+{ 3 }^{ k }) } =\sum _{ k=1 }^{ 11 }{ { 3 }^{ k } } =2\sum_1^{11}(1)+\sum _{ k=1 }^{ 11 }{ { 3 }^{ k } } =2(11)+\sum _{ k=1 }^{ 11 }{ { 3 }^{ k } } =22+\sum _{ k=1 }^{ 11 }{ { 3 }^{ k } } \quad \quad ...(1)\\ \sum _{ k=1 }^{ 11 }{ { 3 }^{ k } } ={ 3 }^{ 1 }+{ 3 }^{ 2 }+{ 3 }^{ 3 }+...{ 3 }^{ 11 }$$
The terms of this sequence $$3, { 3 }^{ 2 },{ 3 }^{ 3, }.........3^{11}$$ forms a G.P.
$$\therefore { S }_{ n }=\dfrac { a({ r }^{ n }-1) }{ r-1 } \\ \Rightarrow { S }_{ n }=\dfrac { 3\left[ { (3) }^{ 11 }-1 \right] }{ 3-1 } \\ \Rightarrow { S }_{ n }=\dfrac { 3 }{ 2 } ({ 3 }^{ 11 }-1)\\ \therefore \sum _{ k=1 }^{ 11 }{ { 3 }^{ k } } =\dfrac { 3 }{ 2 } ({ 3 }^{ 11 }-1)$$
Substituting this value in equation (1), we obtain $$\sum _{ k=1 }^{ 11 }{ { (2+3 }^{ k }) } =22+\dfrac { 3 }{ 2 } ({ 3 }^{ 11 }-1)$$

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