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  • Evaluate summation 11 for k = 1 2 + 3^ k

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

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best_answer

Given :   

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

\sum_{k = 1}^{11} ( 2+ 3 ^k )=\sum _{k=1}^{11}2 +\sum _{k=1}^{11} 3^k

                          =22 +\sum _{k=1}^{11} 3^k...............(1)

\sum _{k=1}^{11} 3^k=3^1+3^2+3^3+....................3^1^1

These terms form GP with a=3 and r=3.

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

S_n=\frac{3(1-3^1^1)}{1-3}

S_n=\frac{3(1-3^1^1)}{-2}

S_n=\frac{3(3^1^1-1)}{2}=\sum _{k=1}^{11} 3^k

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

 

Posted by

seema garhwal

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