according to ramanujan, sum of all natural no. is -1/12 but sum of first N terms is n(n+1)/2...how?
Answers
23 + 43 + 63 + ……… + (2n)3
Even Sum = 23 + 43 + 63 + .... + (2n)3
if we multiply by 23 then
= 23 x (13 + 23 + 32 + .... + (n)3)
= 23 + 43 + 63 + ......... + (2n)3
= 23 (n(n+1)/2)2
= 8(n(n+1))2/4
= 2(n(n+1))2
FOR EXAMPLE:
Example :
Sum of cube of first 4 even numbers = 23 + 43 + 63 + 83
put n = 4 = 2(n(n+1))2
= 2*(4*(4+1))2
= 2(4*5)2
= 2(20)2
= 800
8 + 64 + 256 + 512 = 800
Answer:
Like the above equation, many of Ramanujan’s results involved infinite sums and products that were truly novel and sometimes intimidating even to professional mathematicians. My goal here was to take simple results related to Ramanujan’s work and make them accessible to readers familiar with no more than high school algebra. If, like most people, you find your eyes glazing over when you see infinite nested radicals or infinite continued fractions, I hope this column and the informal explanations below will give you a glimmer of understanding.
Our first question is to prove the following equation involving an infinite nested radical. In 1911, Ramanujan sent the right-hand side of the following equation to a mathematical journal as a puzzle:
3=1+21+31+41+5⋯−−−√−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−√
How can you prove this?
All you need to prove this is the following elementary result:
(x + 1)2 = x2 + 2x + 1 = x(x + 2) + 1
Rearrange the terms to get:
(x + 1)2 = 1 + x(x + 2)
Now if we substitute 2, 3 and 4 for x, we get:
32 = 1 + 2(4), 42 = 1 + 3(5), 52 = 1 + 4(6)…
If we take the square roots of both sides of these equations, we get:
3=1+2(4)−−−−−−−√ (Equation 1)
4=1+3(5)−−−−−−−√ (Equation 2)
5=1+4(6)−−−−−−−√ (Equation 3)
6=1+5(7)−−−−−−−√ (Equation 4)
and so on …
Replacing the (4) in Equation 1 with the right-hand side of Equation 2, we get:
3=1+21+3(5)−−−−−−−√−−−−−−−−−−−−−√
Now replace the (5) above with the right-hand side of Equation 3 to get:
3=1+21+31+4(6)−−−−−−−√−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√
Repeat the same procedure with the next equation:
3=1+21+31+41+5(7)−−−−−−−√−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−√
It is easy to see that we can continue to replace the last integer in parentheses with the right-hand side of the next equation, so the process can be carried out recursively to infinity. So we can replace the 7 with an ellipsis (…) to show that the process can be repeated indefinitely, thus proving the correctness of the equation.
3=1+21+31+41+5⋯−−−√−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−√
Step-by-step explanation: