Intuition The big pictureSigma is just shorthand for "add all these up." Instead of writing
1 + 4 + 9 + 16 + 25 1+4+9+16+25 1 + 4 + 9 + 16 + 25 pattern (k 2 k^2 k 2 ∑ k = 1 5 k 2 \sum_{k=1}^{5}k^2 ∑ k = 1 5 k 2
Standard formulas let us collapse long sums into a closed form.Telescoping — when consecutive terms cancel , a whole sum collapses to just
the first and last survivors. It's the "domino effect run backwards."
Definition Summation notation∑ k = m n a k = a m + a m + 1 + a m + 2 + ⋯ + a n \sum_{k=m}^{n} a_k \;=\; a_m + a_{m+1} + a_{m+2} + \dots + a_n ∑ k = m n a k = a m + a m + 1 + a m + 2 + ⋯ + a n
k k k index of summation (a dummy variable — its name doesn't matter).m m m lower limit (where we start).n n n upper limit (where we stop).a k a_k a k general term (a formula in terms of k k k Number of terms = n − m + 1 = n - m + 1 = n − m + 1
WHY the "+1"? Count k = 2 , 3 , 4 , 5 k=2,3,4,5 k = 2 , 3 , 4 , 5 5 − 2 = 3 5-2=3 5 − 2 = 3 4 4 4 n − m + 1 n-m+1 n − m + 1
WHY? Addition is commutative and associative — you can reorder and regroup a finite
sum freely. Pulling out c c c n n n c c c n n n n c nc n c n n n c c c
∑ ( a k b k ) ≠ ( ∑ a k ) ( ∑ b k ) \sum(a_k b_k) \ne (\sum a_k)(\sum b_k) ∑ ( a k b k ) = ( ∑ a k ) ( ∑ b k ) Why the error feels right: linearity works for + + + Steel-man + fix: test ∑ k = 1 2 k ⋅ k = 1 + 4 = 5 \sum_{k=1}^{2}k\cdot k = 1+4=5 ∑ k = 1 2 k ⋅ k = 1 + 4 = 5 ( ∑ k ) ( ∑ k ) = 3 ⋅ 3 = 9 (\sum k)(\sum k)=3\cdot3=9 ( ∑ k ) ( ∑ k ) = 3 ⋅ 3 = 9 sums and constant factors split.
Intuition Gauss's trick (sum of firstn n n S = 1 + 2 + ⋯ + n S = 1 + 2 + \dots + n S = 1 + 2 + ⋯ + n S = n + ( n − 1 ) + ⋯ + 1 S = n + (n-1) + \dots + 1 S = n + ( n − 1 ) + ⋯ + 1 ( n + 1 ) (n+1) ( n + 1 ) n n n 2 S = n ( n + 1 ) ⇒ ∑ k = 1 n k = n ( n + 1 ) 2 2S = n(n+1) \;\Rightarrow\; \boxed{\sum_{k=1}^{n} k = \frac{n(n+1)}{2}} 2 S = n ( n + 1 ) ⇒ k = 1 ∑ n k = 2 n ( n + 1 )
HOW to derive ∑ k 2 \sum k^2 ∑ k 2
Use the identity ( k + 1 ) 3 − k 3 = 3 k 2 + 3 k + 1 (k+1)^3 - k^3 = 3k^2 + 3k + 1 ( k + 1 ) 3 − k 3 = 3 k 2 + 3 k + 1 k = 1 k=1 k = 1 n n n left side telescopes (see next section) to ( n + 1 ) 3 − 1 (n+1)^3 - 1 ( n + 1 ) 3 − 1 ( n + 1 ) 3 − 1 = 3 ∑ k 2 + 3 ∑ k + ∑ 1 = 3 ∑ k 2 + 3 ⋅ n ( n + 1 ) 2 + n . (n+1)^3 - 1 = 3\sum k^2 + 3\sum k + \sum 1 = 3\sum k^2 + 3\cdot\tfrac{n(n+1)}{2} + n. ( n + 1 ) 3 − 1 = 3 ∑ k 2 + 3 ∑ k + ∑ 1 = 3 ∑ k 2 + 3 ⋅ 2 n ( n + 1 ) + n . ∑ k 2 \sum k^2 ∑ k 2 n ( n + 1 ) ( 2 n + 1 ) 6 \frac{n(n+1)(2n+1)}{6} 6 n ( n + 1 ) ( 2 n + 1 ) Why this trick? Because
the cube-difference knows about k 2 k^2 k 2
Intuition Dominoes that cancelSuppose the term is a difference of a function at two spots : a k = f ( k ) − f ( k + 1 ) a_k = f(k) - f(k+1) a k = f ( k ) − f ( k + 1 ) [ f ( 1 ) − f ( 2 ) ] + [ f ( 2 ) − f ( 3 ) ] + ⋯ + [ f ( n ) − f ( n + 1 ) ] \big[f(1)-f(2)\big]+\big[f(2)-f(3)\big]+\dots+\big[f(n)-f(n+1)\big] [ f ( 1 ) − f ( 2 ) ] + [ f ( 2 ) − f ( 3 ) ] + ⋯ + [ f ( n ) − f ( n + 1 ) ] annihilates . Only f ( 1 ) f(1) f ( 1 ) − f ( n + 1 ) -f(n+1) − f ( n + 1 )
HOW to spot a telescoping sum: the general term is often a fraction whose denominator
factors into consecutive-ish pieces , e.g. 1 k ( k + 1 ) \frac{1}{k(k+1)} k ( k + 1 ) 1 partial fractions to
split it into f ( k ) − f ( k + 1 ) f(k)-f(k+1) f ( k ) − f ( k + 1 )
Worked example 1 — Evaluate∑ k = 1 20 ( 3 k + 2 ) \sum_{k=1}^{20}(3k+2) ∑ k = 1 20 ( 3 k + 2 ) Step 1: Split by linearity: 3 ∑ k = 1 20 k + ∑ k = 1 20 2 3\sum_{k=1}^{20}k + \sum_{k=1}^{20}2 3 ∑ k = 1 20 k + ∑ k = 1 20 2 Why? Sigma distributes over + + + Step 2: ∑ 1 20 k = 20 ⋅ 21 2 = 210 \sum_{1}^{20}k = \frac{20\cdot21}{2}=210 ∑ 1 20 k = 2 20 ⋅ 21 = 210 ∑ 1 20 2 = 20 ⋅ 2 = 40 \sum_{1}^{20}2 = 20\cdot2 = 40 ∑ 1 20 2 = 20 ⋅ 2 = 40 Why? Standard formula + "constant added 20 times."
Step 3: 3 ( 210 ) + 40 = 630 + 40 = 670 3(210) + 40 = 630 + 40 = \mathbf{670} 3 ( 210 ) + 40 = 630 + 40 = 670
Worked example 2 — Telescope∑ k = 1 n 1 k ( k + 1 ) \displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)} k = 1 ∑ n k ( k + 1 ) 1 Step 1 (partial fractions): 1 k ( k + 1 ) = 1 k − 1 k + 1 \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} k ( k + 1 ) 1 = k 1 − k + 1 1 Why? Set 1 k ( k + 1 ) = A k + B k + 1 \frac{1}{k(k+1)}=\frac{A}{k}+\frac{B}{k+1} k ( k + 1 ) 1 = k A + k + 1 B A = 1 , B = − 1 A=1,B=-1 A = 1 , B = − 1 f ( k ) − f ( k + 1 ) f(k)-f(k+1) f ( k ) − f ( k + 1 ) f ( k ) = 1 k f(k)=\tfrac1k f ( k ) = k 1 Step 2: By the telescoping formula, sum = f ( 1 ) − f ( n + 1 ) = 1 − 1 n + 1 = f(1)-f(n+1) = 1 - \frac{1}{n+1} = f ( 1 ) − f ( n + 1 ) = 1 − n + 1 1 Why? All interior terms cancel; only first head and last tail remain.
Step 3: = n n + 1 = \dfrac{n}{n+1} = n + 1 n Check n = 1 n=1 n = 1 1 1 ⋅ 2 = 1 2 \frac{1}{1\cdot2}=\tfrac12 1 ⋅ 2 1 = 2 1 1 2 \frac{1}{2} 2 1
Worked example 3 — Telescope∑ k = 1 n 1 k ( k + 2 ) \displaystyle\sum_{k=1}^{n}\frac{1}{k(k+2)} k = 1 ∑ n k ( k + 2 ) 1 Step 1: 1 k ( k + 2 ) = 1 2 ( 1 k − 1 k + 2 ) \frac{1}{k(k+2)}=\frac12\Big(\frac1k-\frac{1}{k+2}\Big) k ( k + 2 ) 1 = 2 1 ( k 1 − k + 2 1 ) Why? Denominator gap is 2, so cancellation skips one term — expect two survivors at each end.
Step 2: Writing terms, 1 k \frac1k k 1 1 k + 2 \frac{1}{k+2} k + 2 1 1 1 + 1 2 \frac11+\frac12 1 1 + 2 1 − 1 n + 1 − 1 n + 2 -\frac{1}{n+1}-\frac{1}{n+2} − n + 1 1 − n + 2 1 Step 3: Sum = 1 2 ( 1 + 1 2 − 1 n + 1 − 1 n + 2 ) = 1 2 ( 3 2 − 1 n + 1 − 1 n + 2 ) =\frac12\Big(1+\frac12-\frac{1}{n+1}-\frac{1}{n+2}\Big)=\frac12\Big(\frac32-\frac{1}{n+1}-\frac{1}{n+2}\Big) = 2 1 ( 1 + 2 1 − n + 1 1 − n + 2 1 ) = 2 1 ( 2 3 − n + 1 1 − n + 2 1 ) Why two survivors? Because the gap of 2 means each cancellation is "delayed" by one term.
Common mistake Forgetting to re-index after shifting limitsIf you rewrite ∑ k = 1 n 1 k + 1 \sum_{k=1}^{n}\frac{1}{k+1} ∑ k = 1 n k + 1 1 ∑ j = 2 n + 1 1 j \sum_{j=2}^{n+1}\frac1j ∑ j = 2 n + 1 j 1 both
limits when you substitute j = k + 1 j=k+1 j = k + 1 Why the slip feels right: the summand looks simpler, so you
forget the bounds moved with it. Fix: every index shift moves the limits too.
Recall Feynman: explain to a 12-year-oldImagine a long line of dominoes standing up. Sigma just says "count everything in the line."
A telescoping sum is a magic line where each domino is a plus and the very next is the same
number as a minus — so they knock each other out! After all the cancelling, only the first
and the last dominoes are left standing. So instead of adding a hundred things, you just do
"first minus last." That's the whole trick.
Mnemonic Remember telescoping"A collapsing telescope: only the two ends stick out."
The middle slides shut (cancels), leaving f ( start ) − f ( end + 1 ) f(\text{start}) - f(\text{end}+1) f ( start ) − f ( end + 1 )
What does ∑ k = m n a k \sum_{k=m}^{n}a_k ∑ k = m n a k a m + a m + 1 + ⋯ + a n a_m + a_{m+1} + \dots + a_n a m + a m + 1 + ⋯ + a n — add the general term for every integer
k k k from
m m m to
n n n .
How many terms are in ∑ k = m n \sum_{k=m}^{n} ∑ k = m n State ∑ k = 1 n k \sum_{k=1}^{n}k ∑ k = 1 n k n ( n + 1 ) 2 \dfrac{n(n+1)}{2} 2 n ( n + 1 ) (Gauss: pair forwards+backwards,
2 S = n ( n + 1 ) 2S=n(n+1) 2 S = n ( n + 1 ) ).
State ∑ k = 1 n k 2 \sum_{k=1}^{n}k^2 ∑ k = 1 n k 2 n ( n + 1 ) ( 2 n + 1 ) 6 \dfrac{n(n+1)(2n+1)}{6} 6 n ( n + 1 ) ( 2 n + 1 ) .
State ∑ k = 1 n k 3 \sum_{k=1}^{n}k^3 ∑ k = 1 n k 3 [ n ( n + 1 ) 2 ] 2 \left[\dfrac{n(n+1)}{2}\right]^2 [ 2 n ( n + 1 ) ] 2 (square of the sum of integers).
Why does ∑ ( c a k ) = c ∑ a k \sum(c\,a_k)=c\sum a_k ∑ ( c a k ) = c ∑ a k Distributive law:
c c c factors out of every term.
Is ∑ a k b k = ( ∑ a k ) ( ∑ b k ) \sum a_k b_k = (\sum a_k)(\sum b_k) ∑ a k b k = ( ∑ a k ) ( ∑ b k ) No — products don't distribute over Σ (counterexample:
k ⋅ k k\cdot k k ⋅ k ).
General telescoping result? ∑ k = m n [ f ( k ) − f ( k + 1 ) ] = f ( m ) − f ( n + 1 ) \sum_{k=m}^{n}[f(k)-f(k+1)] = f(m)-f(n+1) ∑ k = m n [ f ( k ) − f ( k + 1 )] = f ( m ) − f ( n + 1 ) .
How do you turn 1 k ( k + 1 ) \frac{1}{k(k+1)} k ( k + 1 ) 1 Partial fractions:
1 k − 1 k + 1 \frac1k-\frac1{k+1} k 1 − k + 1 1 .
Evaluate ∑ k = 1 n 1 k ( k + 1 ) \sum_{k=1}^{n}\frac{1}{k(k+1)} ∑ k = 1 n k ( k + 1 ) 1 1 − 1 n + 1 = n n + 1 1-\frac{1}{n+1}=\frac{n}{n+1} 1 − n + 1 1 = n + 1 n .
Why do gap-2 sums like 1 k ( k + 2 ) \frac{1}{k(k+2)} k ( k + 2 ) 1 Cancellation is delayed by one term, so two heads and two tails remain.
Arithmetic Progressions — ∑ ( a + ( k − 1 ) d ) \sum(a+(k-1)d) ∑ ( a + ( k − 1 ) d ) ∑ k \sum k ∑ k Partial Fractions — the key tool to create telescoping structure.Method of Differences — telescoping is its finite-sum incarnation.Mathematical Induction — used to prove the standard Σ formulas rigorously.Convergence of Series — telescoping gives exact partial sums, so limits are easy.Binomial Theorem — expansions are written compactly with Σ.
General term a_k, limits m to n
Gauss forwards+backwards trick
Intuition Hinglish mein samjhoDekho, Sigma (Σ \Sigma Σ 1 + 4 + 9 + 16 1+4+9+16 1 + 4 + 9 + 16 k 2 k^2 k 2 ∑ k 2 \sum k^2 ∑ k 2 k k k n − m + 1 n-m+1 n − m + 1
Sabse mazedaar cheez hai telescoping . Socho dominoes ki line — har term ek + + + − - − pehla aur aakhri bacha rehta hai. Formula: ∑ [ f ( k ) − f ( k + 1 ) ] = f ( 1 ) − f ( n + 1 ) \sum[f(k)-f(k+1)] = f(1)-f(n+1) ∑ [ f ( k ) − f ( k + 1 )] = f ( 1 ) − f ( n + 1 ) 1 k ( k + 1 ) \frac{1}{k(k+1)} k ( k + 1 ) 1 1 k − 1 k + 1 \frac1k-\frac1{k+1} k 1 − k + 1 1 1 − 1 n + 1 1-\frac{1}{n+1} 1 − n + 1 1
Yaad rakho: linearity (+ + + product split nahi hota —
∑ a k b k ≠ ( ∑ a k ) ( ∑ b k ) \sum a_k b_k \ne (\sum a_k)(\sum b_k) ∑ a k b k = ( ∑ a k ) ( ∑ b k ) j = k + 1 j=k+1 j = k + 1
