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

Asked by Pragya Singh | 1 year ago |  66

##### Solution :-

$$\displaystyle\sum_{k=1}^{11} (2+3k)=\displaystyle\sum_{k=1}^{11} (2)+$$

$$\displaystyle\sum_{k=1}^{11} (3k)=22\displaystyle\sum_{k=1}^{11} (3^k)$$ .......(1)

We know that,

$$\displaystyle\sum_{k=1}^{11} (3^k)=3^1+3^2+....+3^{11}$$

This sequence $$3,,3^2,3^3....3^{11}$$ forms a G.P. Therefore,

$$s_n=\dfrac{a(1-r^n)}{1-r}$$

Substituting the values to the above equation we get,

$$s_n=\dfrac{3[(3)^{11}-1]}{3-1}$$

$$s_n=\dfrac{3}{2}(3^{11}-1)$$

Therefore,

$$\displaystyle\sum_{k=1}^{11} (3^k)=\dfrac{3}{2}(3^{11}-1)$$

Substitute this value in equation (1).

$$\displaystyle\sum_{k=1}^{11} (2+3k)=22+\dfrac{3}{2}(3^{11}-1)$$

Answered by Abhisek | 1 year ago

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