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Q. 58

Expert-verifiedFound in: Page 327

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Given a simple proof that $\sum _{k=5}^{n}({a}_{k}+{b}_{k})=\sum _{k=5}^{n}{a}_{k}+\sum _{k=5}^{n}{b}_{k}$

We prove the statement $\sum _{k=5}^{n}({a}_{k}+{b}_{k})=\sum _{k=5}^{n}{a}_{k}+\sum _{k=5}^{n}{b}_{k}$

We are given a relation

$\sum _{k=5}^{n}({a}_{k}+{b}_{k})=\sum _{k=5}^{n}{a}_{k}+\sum _{k=5}^{n}{b}_{k}$. We have to prove this

Consider LHS

We have,

$\sum _{k=5}^{n}({a}_{k}+{b}_{k})\phantom{\rule{0ex}{0ex}}=({a}_{5}+{b}_{5})+({a}_{6}+{b}_{6})+....+({a}_{n}+{b}_{n})\phantom{\rule{0ex}{0ex}}=({a}_{5}+{a}_{6}+...+{a}_{n})+({b}_{5}+{b}_{6}+....+{b}_{5})\phantom{\rule{0ex}{0ex}}=\sum _{k=5}^{n}{a}_{k}+\sum _{k=5}^{n}{b}_{k}\phantom{\rule{0ex}{0ex}}=RHS$

Hence proved

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