" The sum of the series "1+(k)/(3)+(k(k+1))/(3.6)+(k(k+1)(k+2))/(3.6.9)+...

Find the sum of the series sum_(k=1)^(360)((1)/(k sqrt(k+1)+(k+1)sqrt(k)))

Find the sum of the series sum_(k=1)^(360)(1/(ksqrt(k+1)+(k+1)sqrt(k)))

Find the sum of the series sum_(k=1)^(360)(1/(ksqrt(k+1)+(k+1)sqrt(k)))

Statement - 1 : The sum of the series 1+ ( 1+2+4) + ( 4+6+9) + ( 9+ 12+ 16) + ... + ( 361 + 380 + 400) is 8000 Statement-2: sum_(k=1)^(n) [ k^(3) - (k-1)^(3) ] = n^(3) , for any natural number 'n'

Statement 1 : The sum of the series 1 + (1 + 2 + 4)+(4 + 6 + 9) + (9 + 12 + 16)+....+(361 + 380 + 400) is 8000 . Statement 2 : sum_(k = 1)^n (k^3 (k - 1)^3) = n^3 for any natural number n .

Statement 1 The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+"……."+(361+380+400) is 8000. Statement 2 sum_(k=1)^(n)(k^(3)-(k-1)^(3))=n^(3) for any natural number n.

Statement 1 The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+"……."+(361+380+400) is 8000. Statement 2 sum_(k=1)^(n)(k^(3)-(k-1)^(3))=n^(3) for any natural number n.

Statement 1 : The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+….+(361 +380 +400) is 8000 Statement 1: Sigma_(k=1)^(n) (k^3-(k-1)^3)=n^3 , for any natural number n.

A special die is so constructed that the probabilities of throwing 1, 2, 3, 4, 5 and 6 are (1-k)/(6), (1+2k)/(6), (1-k)/(6), (1+k)/(6), (1-2k)/(6) and (1+k)/(6) , respectively. If two such dice are thrown and the probability of getting a sum equal to 9 lies between (1)/(9) and (2)/(9) , then the integral value of k is