# Sum up the following series: $\sum^n_{k=1}\frac{k^2}{2^k}$ [duplicate]

How to find an explicit formula for the term $$\sum^n_{k=1}\frac{k^2}{2^k}$$ Then I discovered that
$$\sum^n_{k=1}\frac{k^2}{2^k}=1^22^n+2^22^{n-1}+3^22^{n-2}+\cdots+n^22^1-\frac{n(n+1)(2n+1)}{6}$$ But how to write $1^22^n+2^22^{n-1}+3^22^{n-2}+\cdots+n^22^1$ into explicit formula?? $$\\$$ A hint would be grateful.

## marked as duplicate by Antonio Vargas, Did sequences-and-series 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 27 '15 at 16:39

• Hint: Differentiate once and twice the series $$\sum_{n=0}^\infty x^n=\frac1{1-x}$$ – Did Oct 27 '15 at 16:34