
Find the sum of the series: 6∑k= 13 · (4)k−1

Answer to Problem 11CYU
Sum of the series is Sn= 4095
Explanation of Solution
Given:
Find the sum of the series: 6∑k= 13 · (4)k−1
Concept Used:
K=1 is the first term and k = 6 is the last term.
Calculation:
Number of terms: n = 6
First term: k=1; 6∑k= 13 · (4)k−1⇒3 · 41−1=3 · 40=3
Last term: k=6; 6∑k= 13 · 4k−1⇒3 · 46−1=3 · 45=3072
Find the common difference:
a1= 3; a6= 3072; find ra6= 3072 = a1 · r5=3 · r5r5=30723 =1024= 45r=4
a1=3; r=4; n=6; Find SnSn= a1(1− rn)1−rSn= 3(1− 46)1−4Sn=3( 1− 46)− 3=4095Sn= 4095
Thus, sum of the series is Sn= 4095
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