Sums and approximation lets tackle the inner sum: 2j-1= 2 (k-1)k (k-1) (k-1) On the first line, were using a standard maneuver: break up the sum of a polynomial into one sum for each term. In this case, we break the sum of 2j-1 into one sum involving j and one sum for the constant. This allows us to apply the standard formulas above, and the rest is simplification. Well use the same trick to finish off the whole pyramid problem, though now we have to sum values of a quadratic polynomial # blocks n+ h pyramid=∑(2k-1)+2∑2j-1 k=1 k-1)+2(k-1) k=1 k2-2>k+ k=1 n(m+5)(m+1 n+ Were done! But let's check a couple easy cases to make sure we made no algebra mis- takes. This formula says that a pyramid of size 1 contains(2. 15+1)/3= 1 block and a pyramid of size 2 contains 2)/3=6 blocks-both of which are correct 2.2 Where do the formulas come from? Sure, we can prove all the summation formulas in Lemma 3 by induction, but where did the expressions on the right come from? How on earth would we even guess an analogous summation formula for, say, fourth powers? Here is systematic way to generate a good guess. Remember that sums are the discrete cousins of integrals. So we might guess that the sum of a degree-k polynomial is a degree (k+ 1) polynomial, just as if we were integrating. Based on this observation, we might� � � � � � � � � � � � � � � � 8 Sums and Approximations let’s tackle the inner sum: k−1 k−1 k−1 2j − 1 = 2 j − 1 j=1 j=1 j=1 (k − 1)k = 2 · − (k − 1) 2 = (k − 1)2 On the first line, we’re using a standard maneuver: break up the sum of a polynomial into one sum for each term. In this case, we break the sum of 2j − 1 into one sum involving j and one sum for the constant. This allows us to apply the standard formulas above, and the rest is simplification. We’ll use the same trick to finish off the whole pyramid problem, though now we have to sum values of a quadratic polynomial: n k−1 # blocks in n­th pyramid = (2k − 1) + 2 · 2j − 1 k=1 j=1 n = (2k − 1) + 2(k − 1)2 k=1 n = 2k2 − 2k + 1 k=1 n n n = 2 k2 − 2 k + 1 k=1 k=1 k=1 1 n(n + 2 )(n + 1) n(n + 1) = 2 · − 2 · + n 3 2 2n3 + n = 3 We’re done! But let’s check a couple easy cases to make sure we made no algebra mis￾takes. This formula says that a pyramid of size 1 contains (2 · 13 + 1)/3 = 1 block and a pyramid of size 2 contains (2 · 23 + 2)/3 = 6 blocks— both of which are correct. 2.2 Where Do the Formulas Come From? Sure, we can prove all the summation formulas in Lemma 3 by induction, but where did the expressions on the right come from? How on earth would we even guess an analogous summation formula for, say, fourth powers? Here is systematic way to generate a good guess. Remember that sums are the discrete cousins of integrals. So we might guess that the sum of a degree­k polynomial is a degree­ (k + 1) polynomial, just as if we were integrating. Based on this observation, we might