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Maths Question 8 – JEE-MAIN 2025

If r=19(r+32r)9Cr=α(32)9β, α,βN, then (α+β)2 is equal to

The sum involves binomial coefficients and powers, suggesting the use of binomial expansion properties.

Step 1: Decompose the Sum✦ Active

Split the given summation into two parts based on the numerator (r+3):

S=r=19(r+32r)9Cr=r=19r2r9Cr+3r=1912r9Cr
Step 2: Apply Binomial Summation Identities○ Expand

Use the identities r=1nrnCrxr=nx(1+x)n1 and r=1nnCrxr=(1+x)n1 with n=9 and x=12.

S1=r=19r9Cr(12)r=912(1+12)91=92(32)8
S2=3r=199Cr(12)r=3[(1+12)91]=3[(32)91]
💡 Teacher's Secret Hint

Remember that the sum for r=0 is excluded in the given problem, so adjust the standard binomial expansion accordingly.

Step 3: Combine and Simplify○ Expand

Substitute S1 and S2 back into the original sum and simplify to match the target form α(32)9β:

S=S1+S2=92(32)8+3[(32)91]

Rewrite (32)8=(32)9(23):

S=92(32)9(23)+3(32)93
S=3(32)9+3(32)93
S=6(32)93

Comparing this with α(32)9β, we get α=6 and β=3. Both are natural numbers. Finally, calculate (α+β)2:

(α+β)2=(6+3)2=92=81
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