1+(1+3)+(1+3+5)+

Prove that the sum of the numbers in any group in the following is the square of an odd integer: (1),(1, 3, 5),(1, 3, 5, 7, 9),... .

Sum of the series 1 (1) + 2(1 + 3) + 3(1+3+5)+4(1+3+5+7)+…..+10(1+3+5+7+…..+19) is equal to

1+ ((1)/(3) + (1)/(3^2) ) + (( 1)/( 3^3) + (1)/( 3^4) + (1)/( 3^5) ) + .... sum of the terms in the n^( th) bracket=

The sum of the series 1/(1*3)+2/(1*3*5)+3/(1*3*5*7)+… is equal to :

(1)/(1!) + (1 + 3)/(2!)*x + (1 + 3 + 5)/(3!)*x^(3) …

The value of 1 (2)/(3) +[16(1)/(2) - { 1 (1)/(10) (3(1)/(3) +5)}] 1/4 "of" 2 (1)/(2) xx 4 (4)/(5) is :