Question

find the sum of the 37th bracket of the following series

( 1 )+ ( 7+7^{2}+7^{3})+ ( 7^{4}+7^{5}+7^{6}+7^{7}+7^{8})+ ( 7^{9}+7^{10}+-------+7^{15})+------
correct answer and explanation is marked as branliest

Answer 1

Answer:

See the starting of each backet is a perfect square (n-1)^2

So the first term of 37th bracket will be

7^(36^2)= 7^1296

So the terms are

7^1296 + 7^1297 + 7^1298 and so on

Take 7^1296 common

7^1296 ( 7^0 + 7^1 + 7^2...)

So it forms a GP and there are (2n-1)terms = 73 terms

Now apply the formula .

[7^1296 ( 7^73-1)]/6