Simplification & Surds · Topic 8 of 8
➿ Telescoping Sums
1 exam question type, fully solvedA long sum like looks like a lot of adding. But each term can be split so that almost everything cancels — like a collapsing telescope.
Split, then watch it cancel
The key split is . Write every term that way and line them up:
Every is killed by the next , every by the next , and so on. Only the first piece and the last piece survive.
Shortcuts: . If the factors jump by a gap (like ), the sum .
➿Telescoping SumWatch the middle cancel
terms n = 5
Each term splits: 1/(k(k+1)) = 1/k − 1/(k+1).
(1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + (1/4 − 1/5) + (1/5 − 1/6)
The inner pieces cancel in pairs, leaving only the very first and very last:
= 1/1 − 1/6
= 5/6
Sum = 5/6
.
📝Practice Questions
Q11/(1×2) + 1/(2×3) + 1/(3×4) + 1/(4×5) equals:
Q21/(1×3) + 1/(3×5) + 1/(5×7) equals:
Q3The sum (2−1)+(3−2)+(4−3)+…+(20−19) equals:
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Real exam questions — Telescoping Sums
1 question types · 3 solved examples from real SSC papersSplit each term into a difference and watch the middle cancel like dominoes — only the ends survive.
How to solve this type
Split each term into a difference using partial fractions — for example 1/(n(n+1)) = 1/n − 1/(n+1). When you write the whole sum out this way, every middle piece is undone by the next term, so only the FIRST piece and the LAST piece survive. Shortcut for fraction sums: if the two factors in each denominator jump by a fixed gap d (like 1×3, 3×5, 5×7 where d = 2), then 1/(first×second) + … = (1/d)(1/first − 1/last). And for any chain of differences, (a₂−a₁) + (a₃−a₂) + … + (aₙ−aₙ₋₁) = aₙ − a₁, because everything between the two ends cancels.
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