3.3.8
Hum closed formulas chahte hain in sums ke liye:
∑ k = 1 n 1 , ∑ k = 1 n k , ∑ k = 1 n k 2 , ∑ k = 1 n k 3 . WHY care? n n closed form se aap ek hi shot mein answer nikal sakte ho, chahe n Sabka master trick hai telescoping / difference method : ek aisi function f ( k ) f ( k + 1 ) − f ( k )
Definition Telescoping sumEk sum k = 1 ∑ n ( f ( k + 1 ) − f ( k ) ) collapse ho jaata hai:
∑ k = 1 n ( f ( k + 1 ) − f ( k ) ) = f ( n + 1 ) − f ( 1 ) . WHY: terms likh ke dekho — ( f ( 2 ) − f ( 1 ) ) + ( f ( 3 ) − f ( 2 ) ) + ⋯ + ( f ( n + 1 ) − f ( n ) ) + −
Ab hum har power ke liye clever f ( k ) choose karenge.
Intuition Gauss ka pairing (dual coding — picture it)Sum ko aage aur peeche likho aur stack karo:
S S = 1 + 2 + ⋯ + ( n − 1 ) + n = n + ( n − 1 ) + ⋯ + 2 + 1 ( n + 1 ) n 2 S = n ( n + 1 )
Scratch se derivation — telescoping with f ( k ) = k 3
Why k 3 Kyunki ( k + 1 ) 3 − k 3 k 2
Worked example Puri derivationStep 1. Difference expand karo.
( k + 1 ) 3 − k 3 = 3 k 2 + 3 k + 1. Why this step? Yeh k 2 k 1
Step 2. Dono sides ko k = 1 → n ∑ k = 1 n [ ( k + 1 ) 3 − k 3 ] = ( n + 1 ) 3 − 1. Why? Interior cubes cancel ho jaate hain; sirf ends bachte hain.
Step 3. RHS split hota hai:
∑ ( 3 k 2 + 3 k + 1 ) = 3 ∑ k 2 + 3 ∑ k + ∑ 1. Why? Σ
Step 4. Known results ∑ k = 2 n ( n + 1 ) ∑ 1 = n ( n + 1 ) 3 − 1 = 3 ∑ k 2 + 3 ⋅ 2 n ( n + 1 ) + n .
Step 5. ∑ k 2 ( n + 1 ) 3 − 1 = n 3 + 3 n 2 + 3 n . 3 ∑ k 2 = n 3 + 3 n 2 + 3 n − 2 3 n ( n + 1 ) − n . n 3 + 3 n 2 + 2 n − 2 3 n 2 + 3 n = 2 2 n 3 + 6 n 2 + 4 n − 3 n 2 − 3 n = 2 2 n 3 + 3 n 2 + n 2 n 3 + 3 n 2 + n = n ( 2 n 2 + 3 n + 1 ) = n ( n + 1 ) ( 2 n + 1 ) 3 ∑ k 2 = 2 n ( n + 1 ) ( 2 n + 1 ) ⇒ ∑ k 2 = 6 n ( n + 1 ) ( 2 n + 1 ) . ✓
Derivation — telescoping with f ( k ) = k 4
Worked example Puri derivationStep 1. ( k + 1 ) 4 − k 4 = 4 k 3 + 6 k 2 + 4 k + 1. Why k 4 Uska difference ek k 3
Step 2. k = 1 → n ( n + 1 ) 4 − 1
Step 3. RHS = 4 ∑ k 3 + 6 ∑ k 2 + 4 ∑ k + n
Step 4. Do known sums plug in karo aur solve karo:
4 ∑ k 3 = ( n + 1 ) 4 − 1 − 6 ⋅ 6 n ( n + 1 ) ( 2 n + 1 ) − 4 ⋅ 2 n ( n + 1 ) − n . n 4 ∑ k 3 = n 4 + 2 n 3 + n 2 = n 2 ( n + 1 ) 2 . ∑ k 3 = 4 n 2 ( n + 1 ) 2 = [ 2 n ( n + 1 ) ] 2 .
Sum
Closed form
Leading behaviour
∑ 1 n ∼ n
∑ k 2 n ( n + 1 ) ∼ 2 1 n 2
∑ k 2 6 n ( n + 1 ) ( 2 n + 1 ) ∼ 3 1 n 3
∑ k 3 [ 2 n ( n + 1 ) ] 2 ∼ 4 1 n 4
Intuition 80/20 — ek hi pattern∑ k p ∼ p + 1 n p + 1 ∫ 0 n x p d x
∑ k = 1 20 ( 3 k 2 − 2 k + 5 ) Split (linearity): 3 ∑ k 2 − 2 ∑ k + 5 ∑ 1 Why? Bade unknown sum ko known building blocks mein tod do.
n = 20 ∑ k 2 = 6 20 ⋅ 21 ⋅ 41 = 2870 ∑ k = 2 20 ⋅ 21 = 210 ∑ 1 = 20 = 3 ( 2870 ) − 2 ( 210 ) + 5 ( 20 ) = 8610 − 420 + 100 = 8290 .
Worked example Shifted start wala sum:∑ k = 5 12 k 2 Trick: ∑ k = 5 12 = ∑ k = 1 12 − ∑ k = 1 4 Why? Formulas k = 1 = 6 12 ⋅ 13 ⋅ 25 − 6 4 ⋅ 5 ⋅ 9 = 650 − 30 = 620 .
Common mistake Classic slips ko steel-man karte hainSlip 1: "∑ k 3 = ( ∑ k ) 2 ∑ k 2 = ( ∑ k ) ? Why it feels right: cube result itna sundar hai ki aap ek pattern expect karte ho. Fix: yeh sirf p = 3 ∑ k 2
Slip 2: ∑ k = 1 n k 2 = 6 n ( n + 1 ) ( 2 n + 1 ) k = 1 shift karna bhool jaana. Fix: ∑ a b = ∑ 1 b − ∑ 1 a − 1
Slip 3: ∑ ( 3 k 2 ) ( ∑ 3 k ) 2 3 ( ∑ k ) 2 Fix: constant linearly bahar aata hai: ∑ 3 k 2 = 3 ∑ k 2
Recall Feynman: 12-saal ke bacche ko samjhaoSocho tum ek aisi seedhi chadh rahe ho jahan har step pichhle se zyada unchi hai. Total height pata karne ke liye normally tum har step add karte. Lekin ek shortcut hai: seedhi ki do copies lado — ek upar jaati hui, ek ulti — aur woh ek perfect rectangle ban jaati hain. Rectangle ke squares gino, aadha karo, ho gaya. Yahi ∑ k edges stack hone par cancel ho jaate hain (telescoping), sirf top aur bottom bachta hai — aur us leftover se hum total padh lete hain. Toh ek million cheezein add karne ki jagah, hum sirf do add karte hain.
Mnemonic Chaar yaad rakho"One n, half-product, sixth-triple, square-the-triangle."
∑ 1 = n ∑ k = 2 n ( n + 1 ) ∑ k 2 = 6 n ( n + 1 ) ( 2 n + 1 ) 2 n + 1 6 ∑ k 3 = ( ∑ k ) 2
∑ k = 1 n 1 n
∑ k = 1 n k 2 n ( n + 1 )
∑ k = 1 n k 2 6 n ( n + 1 ) ( 2 n + 1 )
∑ k = 1 n k 3 [ 2 n ( n + 1 ) ] 2
∑ k 2 f ( k ) = k 3 ( k + 1 ) 3 − k 3 = 3 k 2 + 3 k + 1
∑ k 3 f ( k ) = k 4 ( k + 1 ) 4 − k 4 = 4 k 3 + 6 k 2 + 4 k + 1
Telescoping sum ∑ k = 1 n [ f ( k + 1 ) − f ( k )] f ( n + 1 ) − f ( 1 )
∑ k 3 ∑ k
Standard formulas use karke ∑ k = 5 12 k 2 ∑ 1 12 − ∑ 1 4 = 650 − 30 = 620
Bade n ∑ k p p + 1 n p + 1 ∫ 0 n x p d x
∑ k = 1 20 ( 3 k 2 − 2 k + 5 ) 8290
Telescoping Series — har proof mein use hone wala core cancellation trick.Arithmetic Progressions — ∑ k a = 1 , d = 1 Mathematical Induction — har formula ka alternate proof method.Definite Integrals as Limits of Sums — Riemann sums mein ∑ k , ∑ k 2 Binomial Theorem — differences mein use hone wala ( k + 1 ) p Triangular Numbers — ∑ k ∑ k 3 = ( triangular ) 2
expand k+1 cubed minus k cubed
