This formula describes the multiplication rule for finite summation.(∑k=0nak)∑j=0nbj=∑k=0n∑j=0nakbjThis formula is called Langrange's identity.(∑k=1nakbk)2=(∑k=1nak2)∑k=1nbk2−∑k=1n∑j=K+1n(akbj−ajbk)2Explain the difference between these two summation rules. Why is the product of two sums a double row and the square of the sum is a double row? Why does the square intersect the terms but not the product? Provide proof.

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This formula describes the multiplication rule for finite summation.(∑k=0nak)∑j=0nbj=∑k=0n∑j=0nakbjThis formula is called Langrange&#039;s identity.(∑k=1nakbk)2=(∑k=1nak2)∑k=1nbk2−∑k=1n∑j=K+1n(akbj−ajbk)2Explain the difference between these two summation rules. Why is the product of two sums a double row and the square of the sum is a double row? Why does the square intersect the terms but not the product? Provide proof.

Coraducci0d · Accepted Answer

Step 11) Proof:first thing is like (a0+a1+a2+⋯+an)., (b0+b1+b2+⋯+bn)=(a0b0+a1b0+⋯+a0bn)++(a−[1}b0+⋯+a1bn)+⋯+(anb0+⋯+anbn)Step 22) Proof:for this we have to split each termlets take individual terms first LHS(1) (∑k=1nakbk)2=∑k=1nak2bk2+2∑i=jn−1∑j=i+1naibiajbj(∑k=1nakbk)2=∑k=1nak2bk2+2∑i=1n−1∑j=i+1naibiajbjnow lets take RHS(2) (∑k=1nak2)(∑k=1nbk2)=∑i=1n∑j=1nai2bj2=∑k=1nak2bk2+∑i=1n−1∑j=i+1nai2bj2+∑j=1n−1∑i=j+1nai2bj2and(3) ∑i=1n−1∑j=i+1n(aibj−ajbi)2=∑i=1n−1∑j=i+1nai2bj2+∑j=1n−1∑i=j+1nai2bj2−2∑i=1n−1∑j=i+1naibiajbjsuch that, so from this we can write(∑k=1nakbk)2=(∑k=1nak2)∑k=1nbk2−∑k=1n∑j=k+1n(akbj−ajbk)2

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