Computing $\sum_{k=1}^n k 2^k$ by hand [duplicate]

I'm wondering how to determine the closed form for this sum: $$\sum_{k=1}^n k 2^k$$

I'm aware that it is $2 (2^n n - 2^n + 1)$ through Wolfram|Alpha, but I wonder how this would be done by hand. Searches for "summations of products" yielded no results, so I'm assuming that there isn't an easy way to determine summations that include products of functions.

How would one approach computing this summation without the help of technology?

marked as duplicate by Hans Lundmark, Simply Beautiful Art, Mike Haskel, Jack D'Aurizio summation StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 30 '17 at 14:43

• Compute $\sum_{k=1}^n (k+1) 2^k$ instead. What's the derivative of $\sum_{k=1}^n x^{k+1}$ ? – Gabriel Romon Oct 30 '17 at 14:35
• @labbhattacharjee Broken link :P Use proper link formatting [description here](link here) – Simply Beautiful Art Oct 30 '17 at 14:43