The sum of the following series 1+6+(91^2+2^2+3^2)/7+(121^2+2^2+3^2+4^2)/9+(15(1^2+2^2+…+5^2))/11+.... up to 15 terms, is

Question

The sum of the following series 1+6+(91^2+2^2+3^2)/7+(121^2+2^2+3^2+4^2)/9+(15(1^2+2^2+…+5^2))/11+.... up to 15 terms, is:

MARKS Content Team · Accepted Answer

Correct answer: (D) 7820. t_n=(3n1^2+2^2+3^2+…+n^2)/(2n+1)=(3n×nn+12n+1)/(6 2n+1) =1/2 n^3+n^2 ∴S_15=∑n=115t_n=∑n=1151/2n^3+n^2 =1/2∑n=115n^3+∑n=115n^2 =1/2×15×16^22+(15×16×31)/6 ∑n^3=nn+12^2, ∑n^2=(nn+1n+2)/6 =7200+620 =7820

The sum of the following series $1 + 6 + \frac{9 (1^{2} + 2^{2} + 3^{2})}{7} + \frac{12 (1^{2} + 2^{2} + 3^{2} + 4^{2})}{9} + \frac{15 (1^{2} + 2^{2} + \dots + 5^{2})}{11} + . . . .$ up to $15$ terms, is:

Solution

Relevant Questions

The sum of the following series 1+6+912+22+327+1212+22+32+429+15(12+22+…+52)11+.... up to 15 terms, is:

Solution

Relevant Questions

The sum of the following series $1 + 6 + \frac{9 (1^{2} + 2^{2} + 3^{2})}{7} + \frac{12 (1^{2} + 2^{2} + 3^{2} + 4^{2})}{9} + \frac{15 (1^{2} + 2^{2} + \dots + 5^{2})}{11} + . . . .$ up to $15$ terms, is: