This question was uploaded on 10/08/21 on social media accounts.
Summation of n2/2n from n = 1 to n = ∞ (Infinity)
Procedure:
We consider function of x for general case of summation of xnn2. Then Differentiate f(x) and multiply by x. Then Differentiate it again multiply by x. Then we will get general case in terms of x. Replace x with 1/2 to get required result.\
Step 1:
We consider function of x for general case of summation of xnn2.
Let,
f(x)=\sum_{n=1}^\infty x^n
Consider for |x|<1:
\[f(x)=\sum_{n=1}^\infty x^n = \frac{x}{1-x}\]
Step 2:
Differentiate f(x):
\[f'(x)=\sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}\]
Multiply by x, Let g(x) = xf'(x)
\[g(x)=xf'(x)=\sum_{n=1}^\infty nx^{n} = \frac{x}{(1-x)^2}\]
Step 3:
Differentiate g(x):
\[g'(x)=\sum_{n=1}^\infty n^2x^{n-1} = \frac{(1+x)}{(1-x)^3}\]
Multiply by x, Let h(x) = xg'(x)
\[h(x)=xg'(x)=\sum_{n=1}^\infty n^2x^{n} = \frac{x(1+x)}{(1-x)^3}\]
Step 4:
Now we got equation for general case of summation of xnn2 as a function h(x).
Replace x with 1/2,
We get,
\[h\left(\frac{1}{2}\right)=\sum_{n=1}^\infty\frac{n^2}{2^n} = \frac{\frac{1}{2}\left(1+\frac{1}{2}\right)}{\left(1-\frac{1}{2}\right)^3}\]
\[\Rightarrow h\left(\frac{1}{2}\right)=\sum_{n=1}^\infty\frac{n^2}{2^n} = \frac{\frac{1}{2}\times\frac{3}{2}}{\left(\frac{1}{2}\right)^3}\]
\[\Rightarrow h\left(\frac{1}{2}\right)=\sum_{n=1}^\infty\frac{n^2}{2^n} = \frac{\frac{3}{4}}{\frac{1}{8}}\]
\[\Rightarrow h\left(\frac{1}{2}\right)=\sum_{n=1}^\infty\frac{n^2}{2^n} = 6\]
By this, we got Infinite Summation of n2 /2n = 6
Image Solution:
The question relates to Algebra, Summation
