How do you find the number of terms given #s_n=341# and #1+4+16+64+...#?

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Answer 1 · 2016-11-18T05:57:48

#n=5#

This is a geometric series with first term #a_1=1# and common ratio #r# as

#4/1=16/4=64/16=4#

The sum of such a series for first #n# terms is given by

#S_n=a_1xx(r^n-1)/(r-1)# and as #S_n=341#, #a_1=1# and #r=4# and , we have

#1xx(4^n-1)/(4-1)=341#

or #(4^n-1)/3=341#

or #4^n-1=341xx3=1023#

and #4^n=1023+1=1024=4^5#

Hence, #n=5#