Answer: The sum of first 30 terms of the given sequence is 1890.
Step-by-step explanation: We are given to find the sum of first 30 terms of the following sequence :
\(a_n=4n+1.\)
The first few terms of the above sequence are
\(a_1=5,\\\\a_2=9,\\\\a_3=13,\\\\a_4=17,\\\\a_5=21\\\\\vdots~~~~~\vdots~~~~~\vdots\\\\\)
So, the given sequence is an arithmetic one with first term 5 and common difference
d = 9 - 5 = 13 - 9 = 17 - 13 = . . . = 4.
Therefore, the sum of first 30 terms will be
\(S_{30}\\\\\\=\dfrac{30}{2}\{2a_1+(30-1)d\}\\\\\\=15(2\times5+29\times4)\\\\=15(10+116)\\\\=15\times 126\\\\=1890.\)
Thus, the sum of first 30 terms of the given sequence is 1890.
