3. निम्नलिखित श्रेणियों का n पदों तक या
Find the sum of the follow
(i) 4+25+64+121....
का n पदों तक योग ज्ञात कीजिए:
um of the following series upto n terms
(ii) 2 + 6 + 12 + 20 +....
(iii) 1 +9+ 24 + 46 +.....
(iv) 2 +5+8+11+......
(V) 13+3+3+7+......
जानकीनगा . solve the 5 question
Answers
Answer:
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Explanation:
I did not know that u Explanation or not.... But I am giving
Yes interesting question....
1. 214 as u said to give sum of following
2. 30 as there is difference between them of even number..
EG.. 6-2=4
12-6=6
20-12=8
So next might be 20-30=10
3.a1=1a2=1+8=9a3=9+8+7=24a4=24+8+7+7=46
ak=1+8(k−1)+7(k−1)(k−2)2=72k2−52k
Sum of first n terms
=∑k=1n(72k2−52k)
=72∑k=1nk2−52∑k=1nk
=72⋅16n(n+1)(2n+1)−52⋅12n(n+1)
=112n(n+1)⋅(7(2n+1)−15)
=112n(n+1)(14n−8)
=16n(n+1)(7n−4)
4. 14 is answer as three is difference of all
EG.. 5-2=8
8-5=3
11-8=3
So next,x-11=3
X=11+3=14
5. 13+3+3+7+.......not sure.
Answer:
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