Evaluate sum_{j=1}^{11} (2 + 3^j) | vedclass

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Question

(A) The given expression is $\sum_{j=1}^{11} (2 + 3^j)$.We can split the summation into two parts:$\sum_{j=1}^{11} 2 + \sum_{j=1}^{11} 3^j$.The first

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$22 + \frac{3}{2}(3^{11} - 1)$

Vedclass · Answer

(A) The given expression is $\sum_{j=1}^{11} (2 + 3^j)$.We can split the summation into two parts:$\sum_{j=1}^{11} 2 + \sum_{j=1}^{11} 3^j$.The first part is the sum of $11$ terms of $2$,which is $11 	imes 2 = 22$.The second part is a geometric progression with first term $a = 3$,common ratio $r = 3$,and number of terms $n = 11$.The sum of a geometric progression is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Substituting the values,we get $\frac{3(3^{11} - 1)}{3 - 1} = \frac{3}{2}(3^{11} - 1)$.Adding both parts,the total sum is $22 + \frac{3}{2}(3^{11} - 1)$.

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

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