Question
(JEE Main 2023 PYQ — paraphrased) The sum of the first 10 terms of the series
12+2⋅22+32+2⋅42+52+2⋅62+…
(odd terms are squared, even terms are doubled-squared) is to be found.
Solution — Step by Step
Odd-positioned terms: 12,32,52,72,92 (positions 1, 3, 5, 7, 9 — five terms).
Even-positioned terms: 2⋅22,2⋅42,2⋅62,2⋅82,2⋅102 (positions 2, 4, 6, 8, 10 — five terms).
Sodd=12+32+52+72+92=1+9+25+49+81=165
Or use the formula for sum of squares of odd numbers up to (2n−1)2 with n=5:
∑k=1n(2k−1)2=3n(2n−1)(2n+1)=35⋅9⋅11=165
Seven=2(22+42+62+82+102)=2⋅4⋅(12+22+32+42+52)
=8⋅65⋅6⋅11=8⋅55=440
Total:
S10=165+440=605
Sum of first 10 terms = 605.
Why This Works
When a series has alternating patterns (different formulas for odd and even positions), the cleanest approach is to split it into two sub-series, each with a single pattern, and sum independently.
Standard formulas:
- ∑k=1nk2=n(n+1)(2n+1)/6
- ∑k=1n(2k−1)2=n(2n−1)(2n+1)/3
- ∑k=1n(2k)2=2n(n+1)(2n+1)/3
Memorize these — they appear in JEE Main almost every year, often disguised as “split series” problems.
For “alternating pattern” series, write down the first few terms by hand, separate by index parity, then apply standard formulas. This is faster than trying to fit one formula to the whole series.
Alternative Method
Direct computation: write out all 10 terms and add. 1+8+9+32+25+72+49+128+81+200=605. Same answer, but tedious for 20 or 30 terms — the formula approach scales.
Common Mistake
Students lose count of how many terms are in each sub-series. For 10 total terms with alternating types, there are 5 odd-positioned and 5 even-positioned. For 11 terms, there are 6 odd and 5 even. Miscounting by one shifts the whole answer.